Tangents and Normals

In this chapter, we study the fundamental concepts of tangents and normals to curves. Given a function \( y = f(x) \), its graph represents a continuous curve in the Cartesian plane. At any point on this curve, we can define a tangent, which is a straight line that just touches the curve at that point without crossing it (locally). The normal at that point is a line perpendicular to the tangent.

The fundamental idea relies on the derivative \( \frac{dy}{dx} \). Given a curve \( y = f(x) \), the derivative at a point \( (x_0, y_0) \) gives the slope of the tangent at that point. That is,

\[ m_{\text{tangent}} = f'(x_0). \]

Using this, we can write the equation of the tangent line in its point-slope form:

\[ y - y_0 = f'(x_0) (x - x_0). \]

Since the normal is perpendicular to the tangent, its slope is the negative reciprocal of the tangent's slope:

\[ m_{\text{normal}} = -\frac{1}{f'(x_0)}, \quad \text{provided } f'(x_0) \neq 0. \]

Thus, the equation of the normal at \( (x_0, y_0) \) is:

\[ y - y_0 = -\frac{1}{f'(x_0)} (x - x_0). \]

The mathematical methods covered here serve as essential tools for more advanced topics in calculus, including curvature, envelopes, and optimization problems.

Equation of the Tangent Line

Let \( y = f(x) \) be a differentiable function at \( x = a \), and let \( (a, f(a)) \) be a point on the curve. If the derivative \( f'(a) \) exists and is finite, the equation of the tangent line is:

\[ y - f(a) = f'(a) (x - a). \]

The tangent line is well-defined whenever \( f'(a) \) is finite. However, we must analyze the special case of vertical tangents, which arise when the derivative fails to exist in a finite form.

Vertical Tangents

A tangent is vertical when its slope is undefined due to the derivative tending to \( \pm\infty \). That is, if:

\[ \lim_{x \to a^-} f'(x) = \pm \infty \quad \text{and} \quad \lim_{x \to a^+} f'(x) = \pm \infty, \]

then the tangent is vertical and given by the equation:

\[ x = a. \]

The point \( x = a \) is an inflection point with a vertical tangent, meaning that the concavity changes while the curve remains nearly vertical.

A classic example is:

\[ f(x) = x^{1/3}, \]

which has:

\[ f'(x) = \frac{1}{3} x^{-2/3}. \]

Here, we see that:

\[ \lim_{x \to 0^-} f'(x) = -\infty, \quad \lim_{x \to 0^+} f'(x) = +\infty. \]

Thus, \( x = 0 \) is an inflection point with a vertical tangent.

Horizontal Tangents

A tangent is horizontal when \( f'(a) = 0 \), meaning the curve has a zero slope at \( x = a \). In such cases, the tangent line is given by:

\[ y = f(a). \]

Horizontal tangents typically occur at local maxima, local minima, or stationary points.

We call the point where a tangent line touches a curve the point of tangency. If a curve is given by \( y = f(x) \), and the tangent is drawn at \( x = a \), then the point of tangency is \( (a, f(a)) \). At this point, the curve and the tangent have the same function value and slope.

Example

Find the equation of the tangent to the curve

\[ y = 2x^2 - x + 1 \]

at the point whose abscissa is \( x = 1 \).

Solution:

The equation of the tangent to a curve at a given point is given by

\[ y - y_0 = m (x - x_0), \]

where \( m \) is the slope of the tangent and \( (x_0, y_0) \) is the point of tangency.

To determine the slope, differentiate \( y = 2x^2 - x + 1 \) with respect to \( x \):

\[ \frac{dy}{dx} = 4x - 1. \]

Evaluating at \( x = 1 \),

\[ m = 4(1) - 1 = 3. \]

In the problem we are only provided the abscissa of the point of tangency, we also need to calculate the ordinate of the point to write the equation. The corresponding point on the curve is found by substituting \( x = 1 \) into the given equation:

\[ y = 2(1)^2 - 1 + 1 = 2. \]

Thus, the point of tangency is \( (1,2) \).

Substituting \( x_0 = 1 \), \( y_0 = 2 \), and \( m = 3 \) into the tangent equation,

\[ y - 2 = 3(x - 1). \]

Expanding,

\[ y = 3x - 1. \]

Thus, the required equation of the tangent is

\[ y = 3x - 1. \]

Example

Find the equation of the tangent to the curve

\[ y = \sin x \]

at the point where the abscissa is \( x = \frac{\pi}{6} \).

Solution:

The equation of the tangent to a curve at a given point is given by

\[ y - y_0 = m (x - x_0), \]

where \( m \) is the slope of the tangent, and \( (x_0, y_0) \) is the point of tangency.

To determine the slope, differentiate \( y = \sin x \) with respect to \( x \):

\[ \frac{dy}{dx} = \cos x. \]

Evaluating at \( x = \frac{\pi}{6} \),

\[ m = \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}. \]

The corresponding point on the curve is

\[ y_0 = \sin \frac{\pi}{6} = \frac{1}{2}. \]

Thus, the point of tangency is \( \left( \frac{\pi}{6}, \frac{1}{2} \right) \).

Substituting \( x_0 = \frac{\pi}{6} \), \( y_0 = \frac{1}{2} \), and \( m = \frac{\sqrt{3}}{2} \) into the tangent equation,

\[ y - \frac{1}{2} = \frac{\sqrt{3}}{2} \left( x - \frac{\pi}{6} \right). \]

Expanding,

\[ y = \frac{\sqrt{3}}{2} x - \frac{\sqrt{3} \pi}{12} + \frac{1}{2}. \]

Thus, the required equation of the tangent is

\[ y = \frac{\sqrt{3}}{2} x + \left( \frac{1}{2} - \frac{\sqrt{3} \pi}{12} \right). \]

Implicitly Defined Curve

Find the equation of the tangent to the curve

\[ x + 2y = y^x \]

at the point \( (-1,1) \) on it.

Solution:

Rewriting the given equation in a differentiable form,

\[ x + 2y = e^{x \ln y}. \]

Differentiating both sides with respect to \( x \),

\[ 1 + 2y' = y^x \left( \ln y + \frac{x}{y} y' \right). \]

There is no point in simplifying this further; we simply substitute \( x = -1 \) and \( y = 1 \) to calculate \( y' \).

Since \( \ln 1 = 0 \), the equation reduces to

\[ 1 + 2y' = 1^x \left( 0 + \frac{-1}{1} y' \right), \]

which simplifies to

\[ 1 + 2y' = -y'. \]

Solving for \( y' \),

\[ 3y' = -1 \implies y' = -\frac{1}{3}. \]

Thus, the slope of the tangent at \( (-1,1) \) is \( -\frac{1}{3} \).

The equation of the tangent line is given by

\[ y - y_0 = m (x - x_0). \]

Substituting \( (x_0, y_0) = (-1,1) \) and \( m = -\frac{1}{3} \),

\[ y - 1 = -\frac{1}{3} (x + 1). \]

Expanding,

\[ y = -\frac{1}{3} x + \frac{2}{3}. \]

Thus, the required equation of the tangent is

\[ y = -\frac{1}{3} x + \frac{2}{3}. \]

Finding the Tangent to a Curve with a Given Slope

When we need to find a tangent to a curve that has a given slope, or equivalently, a tangent parallel to a given line, we follow a systematic approach.

Let the curve be given by

\[ y = f(x). \]

We assume the point of tangency to be \( (x_0, y_0) \), which means we need to determine \( x_0 \) and \( y_0 \). Since we have two unknowns, we require two independent equations.

1. First equation: The point \( (x_0, y_0) \) lies on the curve, which means it must satisfy the equation of the curve. That is,

   \[ y_0 = f(x_0). \]

2. Second equation: The derivative \( \frac{dy}{dx} \) gives the slope of the tangent at \( x_0 \), so we equate it to the given slope:

   \[ \left. \frac{dy}{dx} \right|_{(x_0, y_0)} = \text{given slope}. \]

These two equations are then solved simultaneously to find \( x_0 \) and \( y_0 \), giving the required point of tangency. Once \( x_0 \) and \( y_0 \) are determined, the equation of the tangent line is given by

\[ y - y_0 = (\text{given slope}) (x - x_0). \]

Example

Find the equation of the tangent to the curve

\[ y = x^2 - 5x + 5 \]

that is parallel to the line

\[ 2y = 4x + 1. \]

Solution:

Let the point of tangency be \( (x_0, y_0) \). Since this point lies on the curve, it must satisfy

\[ y_0 = x_0^2 - 5x_0 + 5. \]

By differentiating the given curve we obtain the general slope.

\[ \frac{dy}{dx} = 2x - 5. \]

The slope of the tangent at \( (x_0, y_0) \) is given by

\[ \left. \frac{dy}{dx} \right|_{(x_0, y_0)} = 2x_0 - 5. \]

Since the required tangent is parallel to the given line \( 2y = 4x + 1 \), we equate the slopes:

\[ 2x_0 - 5 = 2. \]

Solving for \( x_0 \),

\[ 2x_0 = 7 \implies x_0 = \frac{7}{2}. \]

Substituting \( x_0 = \frac{7}{2} \) into the equation of the curve,

\[ y_0 = \left(\frac{7}{2}\right)^2 - 5\left(\frac{7}{2}\right) + 5 = -\frac{1}{4}. \]

Thus, the point of tangency is \( \left(\frac{7}{2}, -\frac{1}{4}\right) \).

The equation of the tangent line at this point is

\[ y - y_0 = m (x - x_0). \]

Substituting \( m = 2 \),

\[ y + \frac{1}{4} = 2 \left( x - \frac{7}{2} \right). \]

Expanding,

\[ y = 2x - \frac{29}{4}. \]

Thus, the required equation of the tangent is

\[ y = 2x - \frac{29}{4}. \]

Example

Let \( S \) be the set of all values of \( x \) for which the tangent to the curve

\[ y = f(x) = x^3 - x^2 - 2x \]

at \( (x, y) \) is parallel to the line segment joining the points \( (1, f(1)) \) and \( (-1, f(-1)) \). Find \( S \).

Solution:

Let the given curve be

\[ y = f(x) = x^3 - x^2 - 2x. \]

The points \( (1, f(1)) \) and \( (-1, f(-1)) \) lie on this curve. Evaluating,

\[ f(1) = (1)^3 - (1)^2 - 2(1) = 1 - 1 - 2 = -2. \]

\[ f(-1) = (-1)^3 - (-1)^2 - 2(-1) = -1 - 1 + 2 = 0. \]

The slope of the line joining these two points is given by

\[ m = \frac{f(1) - f(-1)}{1 - (-1)} = \frac{-2 - 0}{1 + 1} = \frac{-2}{2} = -1. \]

The equation of the tangent to the curve at \( (x_0, y_0) \) is determined by differentiating \( f(x) \):

\[ \frac{dy}{dx} = 3x^2 - 2x - 2. \]

Since the tangent must be parallel to the line segment, its slope must also be \( -1 \). Setting

\[ 3x^2 - 2x - 2 = -1, \]

we obtain

\[ 3x^2 - 2x - 1 = 0. \]

Factoring,

\[ (x - 1)(3x + 1) = 0. \]

Solving for \( x \),

\[ x = 1, \quad x = -\frac{1}{3}. \]

Thus, the required set is

\[ S = \left\{ -\frac{1}{3}, 1 \right\}. \]

Example

Consider the curve given by

\[ y = x + \sin y. \]

A tangent to this curve at some point \((a,b)\) is parallel to the line joining the points

\[ \left(0, \frac{3}{2} \right) \text{ and } \left(\frac{1}{2}, 2\right). \]

Find \( |b - a| \).

Solution:

Differentiating both sides of the given curve with respect to \( x \),

\[ \frac{dy}{dx} = 1 + \cos y \cdot \frac{dy}{dx}. \]

Rearranging,

\[ \frac{dy}{dx} - \cos y \frac{dy}{dx} = 1 \]

\[ \frac{dy}{dx} (1 - \cos y) = 1 \]

\[ \frac{dy}{dx} = \frac{1}{1 - \cos y}. \]

The slope of the given line joining \(\left(0, \frac{3}{2} \right)\) and \(\left(\frac{1}{2}, 2\right)\) is

\[ m = \frac{2 - \frac{3}{2}}{\frac{1}{2} - 0} = \frac{\frac{1}{2}}{\frac{1}{2}} = 1. \]

Since the tangent at \((a, b)\) is parallel to this line,

\[ \frac{dy}{dx} \Big|_{(a,b)} = 1. \]

Equating with the derivative obtained earlier,

\[ \frac{1}{1 - \cos b} = 1. \]

\[ 1 - \cos b = 1. \]

\[ \cos b = 0. \]

Since \(\cos b = 0\), it follows that

\[ \sin b = \pm 1. \]

Using the equation of the curve at \((a, b)\),

\[ b = a + \sin b. \]

\[ b - a = \sin b. \]

Since \(\sin b = \pm 1\),

\[ |b - a| = 1. \]

Tangent from a Given Point to a Curve

Sometimes, we need to determine tangents from a given external point to a curve. The given point, say \( P(x_0, y_0) \), does not lie on the curve, and we seek to find the equations of all possible tangents from \( P \) to the curve.

For many curves, there can be more than one such tangent. The number of tangents depends on the nature of the curve and the position of the external point. To find the equation of the tangent(s), we follow this procedure:

Let the point of tangency be \( T(x_1, y_1) \). If multiple tangents exist, algebraic calculations will provide multiple values of \( x_1, y_1 \). Since there are two unknowns, \( x_1 \) and \( y_1 \), we need two independent equations to determine them.

1. First equation (Curve Condition): Since \( T(x_1, y_1) \) lies on the curve, it must satisfy the equation of the curve, i.e.,

   \[ y_1 = f(x_1). \]

2. Second equation (Slope Condition): The slope of the tangent at \( T(x_1, y_1) \) is obtained by differentiating the given curve. The derivative at \( x_1 \) gives

   \[ \left. \frac{dy}{dx} \right|_{(x_1, y_1)}. \]

   This slope must be equal to the slope of the line joining \( P(x_0, y_0) \) and \( T(x_1, y_1) \), which is given by the coordinate geometry formula:

   \[ \frac{y_1 - y_0}{x_1 - x_0}. \]

   Equating these slopes, we obtain the second equation:

   \[ \left. \frac{dy}{dx} \right|_{(x_1, y_1)} = \frac{y_1 - y_0}{x_1 - x_0}. \]

These two equations are then solved simultaneously to determine \( x_1 \) and \( y_1 \). Once the point(s) of tangency are found, the equation(s) of the tangent line(s) can be written as:

\[ y - y_1 = \left. \frac{dy}{dx} \right|_{(x_1, y_1)} (x - x_1). \]

This method ensures that all possible tangents from the given external point are determined systematically.

Example

Find the equations of all tangents that can be drawn from the point \( (-1, -2) \) to the curve

\[ y = \frac{x^2}{4}. \]

Solution:

Let \( T(x_1, y_1) \) be the point of tangency of the tangent drawn from \( (-1, -2) \).

Since \( T(x_1, y_1) \) lies on the curve,

\[ y_1 = \frac{x_1^2}{4}. \quad \text{(1)} \]

Differentiating both sides of the equaiton of the curve with respect to \( x \),

\[ \frac{dy}{dx} = \frac{2x}{4} = \frac{x}{2}. \]

Evaluating at \( (x_1, y_1) \),

\[ \left. \frac{dy}{dx} \right|_{(x_1, y_1)} = \frac{x_1}{2}. \]

The slope of the tangent line \( PT \) can also be found using coordinate geometry using point P and T that must be equal to the slope of tangent at T,

\[ \frac{x_1}{2} = \frac{y_1 + 2}{x_1 + 1}. \]

Rearranging,

\[ x_1^2 + x_1 = 2y_1 + 4. \quad \text{(2)} \]

Substituting equation (1) into equation (2),

\[ x_1^2 + x_1 = 2 \times \frac{x_1^2}{4} + 4. \]

\[ x_1^2 + x_1 = \frac{x_1^2}{2} + 4. \]

Simplifying,

\[ x_1^2 + 2x_1 - 8 = 0. \]

Factoring,

\[ (x_1 + 4)(x_1 - 2) = 0. \]

\[ x_1 = -4, \quad x_1 = 2. \]

Substituting these values into equation (1),

For \( x_1 = 2 \),

\[ y_1 = \frac{2^2}{4} = \frac{4}{4} = 1. \]

For \( x_1 = -4 \),

\[ y_1 = \frac{(-4)^2}{4} = \frac{16}{4} = 4. \]

Thus, the points of tangency are \( (2,1) \) and \( (-4,4) \).

The equation of the tangent line is given by

\[ y - y_1 = m (x - x_1). \]

For \( (2,1) \), using \( m = \frac{x_1}{2} = \frac{2}{2} = 1 \),

\[ y - 1 = 1(x - 2). \]

\[ y = x - 1. \]

For \( (-4,4) \), using \( m = \frac{x_1}{2} = \frac{-4}{2} = -2 \),

\[ y - 4 = -2(x + 4). \]

\[ y = -2x - 4. \]

Thus, the equations of the two tangents are

\[ y = x - 1 \quad \text{and} \quad y = -2x - 4. \]

Condition of Tangency

A given line

\[ y = mx + c \]

is a tangent to the curve

\[ y = f(x) \]

if they touch at a single point without crossing. Such problems can be easily handled by following this procedure:

1. Let the point of tangency be \( (x_1, y_1) \).

2. Since the point lies on the curve, we must have

   \[ y_1 = f(x_1). \]

3. Since the point also lies on the given line, we must have

   \[ y_1 = m x_1 + c. \]

4. Since the line is a tangent, its slope must match the slope of the curve at \( (x_1, y_1) \), i.e.,

   \[ m = \frac{dy}{dx} \Big|_{(x_1, y_1)}. \]

Now, from these three equations, we eliminate the unknowns \( x_1 \) and \( y_1 \). The resulting equation will provide the required condition of tangency.

Equation of the Normal Line

Let the equation of the curve be

\[ y = f(x). \]

The normal to the curve at a given point is the straight line perpendicular to the tangent at that point.

General Form of the Normal Equation

Let \( (x_1, y_1) \) be the point of tangency. The slope of the tangent at this point is given by

\[ m_{\text{tangent}} = \left. \frac{dy}{dx} \right|_{(x_1, y_1)}. \]

Since the normal is perpendicular to the tangent, its slope is given by the negative reciprocal of the tangent slope:

\[ m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}}, \quad \text{provided that } m_{\text{tangent}} \neq 0. \]

Thus, the equation of the normal at \( (x_1, y_1) \) is

\[ y - y_1 = -\frac{1}{m_{\text{tangent}}} (x - x_1), \]

where \( m_{\text{tangent}} = f'(x_1) \).

Vertical Normal

A vertical normal occurs when the normal line is perpendicular to the x-axis, meaning it has an equation of the form

\[ x = x_1. \]

For the normal to be vertical, the normal slope \( m_{\text{normal}} \) must be undefined, which happens when

\[ m_{\text{tangent}} = 0. \]

That is, vertical normals exist at points where the tangent is horizontal, i.e.,

\[ \frac{dy}{dx} = 0. \]

At any such point \( (x_1, y_1) \), the equation of the vertical normal is simply

\[ x = x_1. \]

Thus, vertical normals always occur at critical points of the function where the derivative vanishes.

Example

Find the equation of the normal to the curve

\[ y(x - 2)(x - 3) = x + 6 \]

at the point where the curve intersects the \(Y\)-axis. Determine the point through which this normal passes.

Solution:

To find the intersection of the curve with the \(Y\)-axis, substitute \( x = 0 \) into the given equation:

\[ y(0 - 2)(0 - 3) = 0 + 6. \]

\[ y(-2)(-3) = 6. \]

\[ 6y = 6. \]

\[ y = 1. \]

Thus, the curve intersects the \(Y\)-axis at \( (0,1) \).

Rewriting the given equation for \( y \),

\[ y = \frac{x + 6}{(x - 2)(x - 3)}. \]

Differentiating both sides,

\[ \frac{dy}{dx} = \frac{(1)(x - 2)(x - 3) - (x + 6)(x - 3 + x - 2)}{(x - 2)^2 (x - 3)^2}. \]

\[ \frac{dy}{dx} = \frac{(x - 2)(x - 3) - (x + 6)(2x - 5)}{(x - 2)^2 (x - 3)^2}. \]

Evaluating at \( (0,1) \),

\[ \left( \frac{dy}{dx} \right)_{(0,1)} = \frac{36}{36} = 1. \]

Since the normal is perpendicular to the tangent, its slope is

\[ m_{\text{normal}} = -\frac{1}{1} = -1. \]

Thus, the equation of the normal at \( (0,1) \) is

\[ y - 1 = -1(x - 0). \]

\[ x + y - 1 = 0. \]

To determine a point through which this normal passes, rewrite the equation as

\[ y = 1 - x. \]

Setting \( x = \frac{1}{2} \),

\[ y = 1 - \frac{1}{2} = \frac{1}{2}. \]

Thus, the normal passes through \( \left( \frac{1}{2}, \frac{1}{2} \right) \).

Example

Consider the function

\[ f(x) = \tan^{-1} \left( \sqrt{\frac{1 + \sin x}{1 - \sin x}} \right), \quad x \in \left(0, \frac{\pi}{2} \right). \]

Find the equation of the normal to the curve \( y = f(x) \) at \( x = \frac{\pi}{6} \).

Solution:

First, we simplify the given expression of the function:

\[ f(x) = \tan^{-1} \left( \sqrt{\frac{1 + \sin x}{1 - \sin x}} \right). \]

Using the identities

\[ 1 + \sin x = \left( \cos \frac{x}{2} + \sin \frac{x}{2} \right)^2, \quad 1 - \sin x = \left( \cos \frac{x}{2} - \sin \frac{x}{2} \right)^2, \]

we rewrite \( f(x) \) as

\[ f(x) = \tan^{-1} \left( \sqrt{\frac{\left( \cos \frac{x}{2} + \sin \frac{x}{2} \right)^2}{\left( \cos \frac{x}{2} - \sin \frac{x}{2} \right)^2}} \right). \]

Since \( \sqrt{z^2} = |z| \), we obtain

\[ f(x) = \tan^{-1} \left| \frac{\cos \frac{x}{2} + \sin \frac{x}{2}}{\cos \frac{x}{2} - \sin \frac{x}{2}} \right|. \]

Now, since \( x \in (0, \pi/2) \), it follows that \( x/2 \in (0, \pi/4) \). In this interval, we have \( \cos \frac{x}{2} > \sin \frac{x}{2} \), so the modulus sign can be removed, giving

\[ f(x) = \tan^{-1} \left( \frac{\cos \frac{x}{2} + \sin \frac{x}{2}}{\cos \frac{x}{2} - \sin \frac{x}{2}} \right). \]

Dividing numerator and denominator by \( \cos \frac{x}{2} \),

\[ f(x) = \tan^{-1} \left( \frac{1 + \tan \frac{x}{2}}{1 - \tan \frac{x}{2}} \right). \]

Using the identity

\[ \frac{1 + \tan A}{1 - \tan A} = \tan(\frac{\pi}{4} + A), \]

we simplify further to

\[ f(x) = \tan^{-1} \tan \left( \frac{\pi}{4} + \frac{x}{2} \right) = \frac{\pi}{4} + \frac{x}{2}. \]

Differentiating both sides,

\[ f'(x) = \frac{1}{2}. \]

Evaluating at \( x = \frac{\pi}{6} \),

\[ f'\left(\frac{\pi}{6}\right) = \frac{1}{2}. \]

Since the normal is perpendicular to the tangent, its slope is

\[ m_{\text{normal}} = -2. \]

Using the point-slope form of the equation of the normal,

\[ y - f\left(\frac{\pi}{6} \right) = -2 \left(x - \frac{\pi}{6} \right). \]

Since

\[ f\left(\frac{\pi}{6}\right) = \frac{\pi}{4} + \frac{\pi}{12} = \frac{\pi}{3}, \]

the equation of the normal becomes

\[ y - \frac{\pi}{3} = -2 \left( x - \frac{\pi}{6} \right). \]

Expanding,

\[ y = -2x + \frac{\pi}{3} + \frac{\pi}{3} = -2x + \frac{2\pi}{3}. \]

Setting \( x = 0 \) to find where the normal intersects the \( Y \)-axis,

\[ y = -2(0) + \frac{2\pi}{3} = \frac{2\pi}{3}. \]

Thus, the normal passes through the point \( \left( 0, \frac{2\pi}{3} \right) \).

Example

Find the equation of the normal to the curve

\[ x^2 + 2xy - 3y^2 = 0 \]

at the point \( (1,1) \).

Solution:

Differentiating both sides implicitly,

\[ 2x + 2x \frac{dy}{dx} + 2y - 6y \frac{dy}{dx} = 0. \]

Rearranging,

\[ (2x - 6y) \frac{dy}{dx} = - (2x + 2y). \]

\[ \frac{dy}{dx} = \frac{- (2x + 2y)}{2x - 6y}. \]

Evaluating at \( (1,1) \),

\[ \left( \frac{dy}{dx} \right)_{(1,1)} = \frac{- (2(1) + 2(1))}{2(1) - 6(1)} = \frac{-4}{-4} = 1. \]

Since the normal is perpendicular to the tangent, its slope is

\[ m_{\text{normal}} = -1. \]

Using the point-slope form,

\[ y - 1 = -1(x - 1). \]

\[ x + y - 2 = 0. \]

Thus, the equation of the normal is

\[ x + y - 2 = 0. \]

Example

Find the equation of the normal to the curve

\[ y = \left| x^2 - 5|x| + 6 \right| \]

at the point whose abscissa is \( x = -1 \).

Solution:

In the small neighborhood around \( x = -1 \), we observe that \( x < 0 \), so \( |x| = -x \). Substituting this into the given function,

\[ y = \left| x^2 - 5(-x) + 6 \right|. \]

\[ = \left| x^2 + 5x + 6 \right|. \]

Factoring the quadratic expression inside the absolute value,

\[ y = \left| (x + 2)(x + 3) \right|. \]

Now, in the small neighborhood around \( x = -1 \), both \( x + 2 \) and \( x + 3 \) are positive, so their product \( (x + 2)(x + 3) \) is positive. This allows us to remove the absolute value, giving

\[ y = (x + 2)(x + 3). \]

Differentiating both sides,

\[ \frac{dy}{dx} = (x + 2) \cdot \frac{d}{dx}(x + 3) + (x + 3) \cdot \frac{d}{dx}(x + 2). \]

\[ = (x + 2)(1) + (x + 3)(1). \]

\[ = (x + 2) + (x + 3). \]

\[ = 2x + 5. \]

Evaluating at \( x = -1 \),

\[ \left( \frac{dy}{dx} \right)_{x=-1} = 2(-1) + 5 = 3. \]

Since the normal is perpendicular to the tangent, its slope is

\[ m_{\text{normal}} = -\frac{1}{3}. \]

The corresponding point on the curve is

\[ y = (-1 + 2)(-1 + 3) = (1)(2) = 2. \]

Using the point-slope form,

\[ y - 2 = -\frac{1}{3} (x + 1). \]

\[ 3(y - 2) = - (x + 1). \]

\[ x + 3y - 7 = 0. \]

Thus, the equation of the normal is

\[ x + 3y - 7 = 0. \]

Tangents and Normals to a Parametric Curve

When the equation of a curve is given in parametric form as

\[ x = f(\theta), \quad y = g(\theta), \]

the tangent at \( \theta \) refers to the tangent line at the point \( (f(\theta), g(\theta)) \), and the normal at \( \theta \) refers to the normal line at the same point.

Equation of the Tangent

Differentiating both equations with respect to \( \theta \),

\[ \frac{dx}{d\theta} = f'(\theta), \quad \frac{dy}{d\theta} = g'(\theta). \]

The slope of the tangent is

\[ \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{g'(\theta)}{f'(\theta)}. \]

Using the point-slope form, the equation of the tangent at \( \theta \) is

\[ y - g(\theta) = \frac{g'(\theta)}{f'(\theta)} (x - f(\theta)). \]

Equation of the Normal

Since the normal is perpendicular to the tangent, its slope is

\[ m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}} = -\frac{f'(\theta)}{g'(\theta)}. \]

Thus, the equation of the normal at \( \theta \) is

\[ y - g(\theta) = -\frac{f'(\theta)}{g'(\theta)} (x - f(\theta)). \]

Example

Prove that the normal to the curve

\[ x = a (\cos\theta + \theta \sin\theta), \quad y = a (\sin\theta - \theta \cos\theta) \]

at any point \( \theta \) is at a constant distance from the origin.

Solution:

Differentiating both parametric equations with respect to \( \theta \),

\[ \frac{dx}{d\theta} = a (-\sin\theta + \sin\theta + \theta \cos\theta) = a \theta \cos\theta, \]

\[ \frac{dy}{d\theta} = a (\cos\theta - \cos\theta + \theta \sin\theta) = a \theta \sin\theta. \]

The slope of the tangent is

\[ \frac{dy}{dx} = \frac{a \theta \sin\theta}{a \theta \cos\theta} = \tan\theta. \]

Since the normal is perpendicular to the tangent, its slope is

\[ m_{\text{normal}} = -\cot\theta. \]

Using the point-slope form, the equation of the normal at \( (x, y) \) is

\[ y - a (\sin\theta - \theta \cos\theta) = -\cot\theta \left[ x - a (\cos\theta + \theta \sin\theta) \right]. \]

Expanding the right-hand side,

\[ y - a \sin\theta + a \theta \cos\theta = -\cot\theta \cdot x + a \cot\theta (\cos\theta + \theta \sin\theta). \]

Rewriting \( \cot\theta = \frac{\cos\theta}{\sin\theta} \),

\[ y - a \sin\theta + a \theta \cos\theta = -\frac{\cos\theta}{\sin\theta} x + a \frac{\cos\theta}{\sin\theta} (\cos\theta + \theta \sin\theta). \]

Multiplying both sides by \( \sin\theta \) to clear the fraction,

\[ y \sin\theta - a \sin^2\theta + a \theta \cos\theta \sin\theta = -x \cos\theta + a \cos^2\theta + a \theta \cos\theta \sin\theta. \]

Rearranging terms,

\[ y \sin\theta + x \cos\theta = a (\cos^2\theta + \sin^2\theta) = a. \]

Thus, the equation of the normal simplifies to

\[ y \sin\theta + x \cos\theta = a. \]

The perpendicular distance of this line from the origin is given by

\[ \frac{|0 \cdot \cos\theta + 0 \cdot \sin\theta - a|}{\sqrt{\cos^2\theta + \sin^2\theta}}. \]

Since \( \cos^2\theta + \sin^2\theta = 1 \),

\[ \text{Distance} = \frac{| -a |}{1} = a. \]

Thus, the normal to the curve is always at a constant distance \( a \) from the origin.

Example

Prove that the sum of the intercepts of the tangent at any point on the curve

\[ \sqrt{x} + \sqrt{y} = \sqrt{a} \]

on the coordinate axes is constant.

Solution:

Since the given curve involves square roots, it is defined for \( 0 \leq x \leq a \) and \( 0 \leq y \leq a \). To simplify differentiation, we express the curve in parametric form. Let

\[ x = a \cos^4\theta, \quad y = a \sin^4\theta. \]

Differentiating both sides with respect to \( \theta \),

\[ \frac{dx}{d\theta} = -4a \cos^3\theta \sin\theta, \]

\[ \frac{dy}{d\theta} = 4a \sin^3\theta \cos\theta. \]

The slope of the tangent is

\[ \frac{dy}{dx} = \frac{4a \sin^3\theta \cos\theta}{-4a \cos^3\theta \sin\theta} = -\frac{\sin^2\theta}{\cos^2\theta}. \]

The equation of the tangent at \( (\tilde{x}, \tilde{y}) = (a\cos^4\theta, a\sin^4\theta) \) is

\[ y - a\sin^4\theta = -\frac{\sin^2\theta}{\cos^2\theta} (x - a\cos^4\theta). \]

Cross-multiplying by \( \cos^2\theta \),

\[ (y - a\sin^4\theta) \cos^2\theta = -\sin^2\theta (x - a\cos^4\theta). \]

Expanding,

\[ y \cos^2\theta - a \sin^4\theta \cos^2\theta = -x \sin^2\theta + a \sin^2\theta \cos^4\theta. \]

Rearranging,

\[ x \sin^2\theta + y \cos^2\theta = a (\sin^2\theta \cos^4\theta + \sin^4\theta \cos^2\theta). \]

Factoring out \( a \sin^2\theta \cos^2\theta \),

\[ x \sin^2\theta + y \cos^2\theta = a \sin^2\theta \cos^2\theta (\cos^2\theta + \sin^2\theta). \]

Since \( \cos^2\theta + \sin^2\theta = 1 \),

\[ x \sin^2\theta + y \cos^2\theta = a \sin^2\theta \cos^2\theta. \]

Finding the Intercepts

For the x-intercept, set \( y = 0 \),

\[ x \sin^2\theta = a \sin^2\theta \cos^2\theta. \]

If \( \sin^2\theta \neq 0 \), dividing by \( \sin^2\theta \),

\[ x = a \cos^2\theta. \]

For the y-intercept, set \( x = 0 \),

\[ y \cos^2\theta = a \sin^2\theta \cos^2\theta. \]

If \( \cos^2\theta \neq 0 \), dividing by \( \cos^2\theta \),

\[ y = a \sin^2\theta. \]

Summing the intercepts,

\[ x_{\text{int}} + y_{\text{int}} = a \cos^2\theta + a \sin^2\theta. \]

Using \( \cos^2\theta + \sin^2\theta = 1 \),

\[ x_{\text{int}} + y_{\text{int}} = a(1) = a. \]

Since this sum is independent of \( \theta \), it remains constant.

Angle of Intersection of Two Curves

If two curves intersect at a point \( P(x_I, y_I) \), the angle of intersection of the two curves at that point is defined as the angle between their respective tangents at \( P \).

Let the given curves be

\[ F(x, y) = 0 \quad \text{and} \quad G(x, y) = 0. \]

To determine the angle between the two curves at \( P(x_I, y_I) \), the slopes of their respective tangents must be found. Differentiating both implicitly and solving for \( \frac{dy}{dx} \), the slopes of the tangents to the curves are

\[ m_1 = \frac{dy}{dx} \Big|_{(x_I, y_I)} \quad \text{for} \quad F(x, y) = 0, \]

\[ m_2 = \frac{dy}{dx} \Big|_{(x_I, y_I)} \quad \text{for} \quad G(x, y) = 0. \]

The acute angle \( \theta \) between the two tangents is given by

\[ \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|. \]

Special Cases

• Orthogonal Intersection: If the curves intersect at \( P(x_I, y_I) \) such that their tangents are perpendicular, then

  \[ m_1 m_2 = -1. \]

  In this case, \( \theta = 90^\circ \), meaning the curves meet at a right angle.

  

• Curves Touching Each Other: If the curves intersect at \( P(x_I, y_I) \) such that

  \[ m_1 = m_2, \]

  then the two curves share the same tangent at \( P(x_I, y_I) \), meaning they touch each other without forming a distinct angle. This implies that the two curves have a common tangent at the point of intersection.

  

Example

Find the angle of intersection of the curves

\[ y = 10 - x^2 \]

and

\[ y = 2 + x^2 \]

at their points of intersection.

Solution:

The points of intersection are obtained by solving

\[ 10 - x^2 = 2 + x^2. \]

Rearranging,

\[ 10 - 2 = x^2 + x^2. \]

\[ 8 = 2x^2. \]

\[ x^2 = 4 \implies x = \pm 2. \]

Substituting \( x = 2 \) into either equation,

\[ y = 10 - 4 = 6. \]

Similarly, for \( x = -2 \),

\[ y = 10 - 4 = 6. \]

Thus, the curves intersect at \( (2,6) \) and \( (-2,6) \).

To find the angle of intersection, compute the slopes of the tangents to both curves at these points. Differentiating the given equations,

\[ \frac{dy}{dx} = -2x \quad \text{for} \quad y = 10 - x^2, \]

\[ \frac{dy}{dx} = 2x \quad \text{for} \quad y = 2 + x^2. \]

At \( (2,6) \),

\[ m_1 = -2(2) = -4, \quad m_2 = 2(2) = 4. \]

The angle \( \theta \) between the tangents is given by

\[ \tan \theta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right|. \]

Substituting values,

\[ \tan \theta = \left| \frac{4 - (-4)}{1 + (-4)(4)} \right| = \left| \frac{4 + 4}{1 - 16} \right|. \]

\[ = \left| \frac{8}{-15} \right| = \frac{8}{15}. \]

Similarly, at \( (-2,6) \),

\[ m_1 = -2(-2) = 4, \quad m_2 = 2(-2) = -4. \]

\[ \tan \theta = \left| \frac{-4 - 4}{1 + (4)(-4)} \right| = \left| \frac{-8}{1 - 16} \right|. \]

\[ = \left| \frac{-8}{-15} \right| = \frac{8}{15}. \]

Thus, in both cases,

\[ \tan \theta = \frac{8}{15}. \]

Use symmetry to reduce calculations

Always use symmetry to avoid extraneous calculations. In the above example, both curves intersect at two distinct points. At both points, the angle of intersection is the same. The reason is the same symmetry in both curves—both are symmetrical about the \( y \)-axis.

To check this symmetry, replace \( x \) with \( -x \) in both equations:

For \( y = 10 - x^2 \), replacing \( x \) with \( -x \) gives

\[ y = 10 - (-x)^2 = 10 - x^2. \]

Since the equation remains unchanged, the curve is symmetric about the \( y \)-axis.

For \( y = 2 + x^2 \), replacing \( x \) with \( -x \) gives

\[ y = 2 + (-x)^2 = 2 + x^2. \]

Again, the equation remains unchanged, confirming symmetry about the \( y \)-axis.

Since both curves are symmetric about the \( y \)-axis, their tangents at \( (2,6) \) and \( (-2,6) \) must form the same angle at both intersection points. This symmetry eliminates the need for redundant calculations at both points.

Thus, if we identify symmetry before calculations, we can use it to avoid extra work and simplify computations efficiently.

Example

Find the angle of intersection of the curves

\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]

and

\[ x^2 + y^2 = ab \]

given that \( a > b \).

Solution:

To determine the points of intersection, express \( y^2 \) from the second equation as

\[ y^2 = ab - x^2. \]

Substituting this into the first equation,

\[ \frac{x^2}{a^2} + \frac{ab - x^2}{b^2} = 1. \]

Multiplying throughout by \( a^2 b^2 \),

\[ b^2 x^2 + a^2 (ab - x^2) = a^2 b^2. \]

\[ b^2 x^2 + a^3 b - a^2 x^2 = a^2 b^2. \]

Rearranging,

\[ (b^2 - a^2)x^2 = a^2 b (b - a). \]

Solving for \( x^2 \),

\[ x^2 = \frac{a^2 b}{a + b} \implies x = \pm \frac{\sqrt{b} a}{\sqrt{a + b}} \]

Using \( y^2 = ab - x^2 \),

\[ y^2 = ab - \frac{a^2 b}{a + b}. \]

\[ = \frac{ab(a + b) - a^2 b}{a + b} \implies y = \pm \frac{\sqrt{a} b}{\sqrt{a + b}}. \]

Thus, the points of intersection are

\[ \left( \pm \frac{a\sqrt{b}}{\sqrt{a + b}}, \pm \frac{\sqrt{a}b}{\sqrt{a + b}} \right). \]

Since, both curves are symmetrical around x-axis and y-axis. There are four intersection points, but the angle of intersection at each intersection point is the same. So we need to just consider one point \( \left( \frac{a\sqrt{b}}{\sqrt{a + b}}, \frac{\sqrt{a}b}{\sqrt{a + b}} \right). \).

To find the angle of intersection, differentiate both curves. Differentiating the first equation implicitly,

\[ \frac{d}{dx} \left( \frac{x^2}{a^2} + \frac{y^2}{b^2} \right) = \frac{d}{dx} (1). \]

\[ \frac{2x}{a^2} + \frac{2y}{b^2} \frac{dy}{dx} = 0. \]

Solving for \( \frac{dy}{dx} \),

\[ \frac{dy}{dx} = -\frac{b^2 x}{a^2 y}. \]

For the second equation,

\[ \frac{d}{dx} (x^2 + y^2) = \frac{d}{dx} (ab). \]

\[ 2x + 2y \frac{dy}{dx} = 0. \]

Solving for \( \frac{dy}{dx} \),

\[ \frac{dy}{dx} = -\frac{x}{y}. \]

Let the slopes of the tangents to the first and second curves be \( m_1 \) and \( m_2 \), respectively:

\[ m_1 = -\frac{b^2 x}{a^2 y}, \quad m_2 = -\frac{x}{y}. \]

The angle \( \theta \) between the two tangents is given by

\[ \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|. \]

Substituting values,

\[ \tan \theta = \left| \frac{-\frac{b^2 x}{a^2 y} + \frac{x}{y}}{1 + \left(-\frac{b^2 x}{a^2 y}\right) \left(-\frac{x}{y}\right)} \right|. \]

\[ = \left| \frac{x}{y} \left( 1 - \frac{b^2}{a^2} \right) \bigg/ \left( 1 + \frac{b^2 x^2}{a^2 y^2} \right) \right|. \]

Since \( x = \frac{a\sqrt{b}}{\sqrt{a + b}} \) and \( y = \frac{\sqrt{a}b}{\sqrt{a + b}} \), \(\frac{x}{y} = \sqrt{\frac{a}{b}}\)

\[ \tan \theta = \left| \sqrt{\frac{a}{b}} \left( 1 - \frac{b^2}{a^2} \right) \bigg/ \left( 1 + \frac{b^2}{a^2} \cdot \frac{a}{b} \right) \right|. \]

\[ = \left| \sqrt{\frac{a}{b}} \cdot \frac{a^2 - b^2}{a^2} \bigg/ \left( 1 + \frac{b}{a} \right) \right|. \]

\[ = \left| \sqrt{\frac{a}{b}} \cdot \frac{a^2 - b^2}{a^2} \bigg/ \frac{a + b}{a} \right|. \]

\[ = \left| \frac{a - b}{\sqrt{ab}} \right| = \left| \frac{a - b}{\sqrt{ab}} \right|. \]

as \(a>b\)

Thus, the angle of intersection is

\[ \theta = \tan^{-1} \left( \frac{a - b}{\sqrt{ab}} \right). \]

Example

If the curves

\[ y^2 = 6x \]

and

\[ 9x^2 + by^2 = 16 \]

intersect at right angles, find the value of \( b \).

Solution:

Let the point of intersection be \( (x_1, y_1) \). Now, we have three unknowns: \( x_1 \), \( y_1 \), and \( b \). To determine \( b \), we form three equations and eliminate \( x_1 \) and \( y_1 \) to obtain the required value.

1. Since \( (x_1, y_1) \) lies on the first curve,

   \[ y_1^2 = 6x_1. \]

2. Since \( (x_1, y_1) \) lies on the second curve,

   \[ 9x_1^2 + b y_1^2 = 16. \]

3. Since the curves intersect at right angles, the product of their slopes must satisfy

   \[ m_1 m_2 = -1. \]

Differentiating the first equation implicitly,

\[ 2y \frac{dy}{dx} = 6. \]

Solving for \( \frac{dy}{dx} \),

\[ \frac{dy}{dx} = \frac{3}{y}. \]

Thus, the slope of the tangent to the first curve at \( (x_1, y_1) \) is

\[ m_1 = \frac{3}{y_1}. \]

Differentiating the second equation implicitly,

\[ 18x + 2by \frac{dy}{dx} = 0. \]

Solving for \( \frac{dy}{dx} \),

\[ \frac{dy}{dx} = \frac{-9x}{by}. \]

Thus, the slope of the tangent to the second curve at \( (x_1, y_1) \) is

\[ m_2 = \frac{-9x_1}{b y_1}. \]

Using the condition of perpendicular tangents,

\[ m_1 m_2 = -1. \]

Substituting the values of \( m_1 \) and \( m_2 \),

\[ \frac{3}{y_1} \cdot \frac{-9x_1}{b y_1} = -1. \]

\[ \frac{-27 x_1}{b y_1^2} = -1. \]

\[ \frac{27 x_1}{b y_1^2} = 1. \]

Using \( y_1^2 = 6x_1 \) from the first equation,

\[ \frac{27 x_1}{b (6x_1)} = 1. \]

\[ \frac{27}{6b} = 1. \]

\[ b = \frac{9}{2}. \]

Thus, the required value of \( b \) is

\[ b = \frac{9}{2}. \]

Example

Prove that if the curves

\[ ax^2 + by^2 = 1 \quad \text{(1)} \]

and

\[ cx^2 + dy^2 = 1 \quad \text{(2)} \]

intersect each other orthogonally, then

\[ \frac{1}{a} - \frac{1}{b} = \frac{1}{c} - \frac{1}{d}. \]

Solution:

Let the point of intersection be \( (x_0, y_0) \). Since this point lies on both curves, it must satisfy equations (1) and (2):

\[ a x_0^2 + b y_0^2 = 1, \quad \text{(3)} \]

\[ c x_0^2 + d y_0^2 = 1. \quad \text{(4)} \]

These two equations provide a system in terms of \( x_0^2 \) and \( y_0^2 \). Instead of solving explicitly for \( x_0^2 \) and \( y_0^2 \), we proceed with differentiation.

Differentiating equation (1) implicitly,

\[ 2a x + 2b y \frac{dy}{dx} = 0. \]

Solving for \( \frac{dy}{dx} \),

\[ \frac{dy}{dx} = -\frac{a x}{b y}. \]

Similarly, differentiating equation (2),

\[ 2c x + 2d y \frac{dy}{dx} = 0. \]

Solving for \( \frac{dy}{dx} \),

\[ \frac{dy}{dx} = -\frac{c x}{d y}. \]

Since the curves intersect orthogonally, the tangents must be perpendicular, so

\[ m_1 m_2 = -1. \]

Substituting the values of \( m_1 \) and \( m_2 \),

\[ \left( -\frac{a x_0}{b y_0} \right) \cdot \left( -\frac{c x_0}{d y_0} \right) = -1. \]

\[ \frac{x_0^2}{y_0^2} \cdot \frac{ac}{bd} = -1. \]

\[ \frac{x_0^2}{y_0^2} = -\frac{bd}{ac}. \quad \text{(5)} \]

Now, subtracting equation (4) from equation (3):

\[ a x_0^2 + b y_0^2 - (c x_0^2 + d y_0^2) = 0. \]

\[ (a - c)x_0^2 + (b - d)y_0^2 = 0. \]

Rearranging,

\[ (a - c)x_0^2 = (d - b)y_0^2. \]

Dividing by \( y_0^2 \),

\[ (a - c) \frac{x_0^2}{y_0^2} = d - b. \]

Substituting (5),

\[ (a - c) \left(-\frac{bd}{ac} \right) = d - b. \]

\[ \frac{a - c}{ac} = \frac{b - d}{bd}. \]

Rearranging,

\[ \frac{1}{a} - \frac{1}{c} = \frac{1}{d} - \frac{1}{b}. \]

Thus,

\[ \frac{1}{a} - \frac{1}{b} = \frac{1}{c} - \frac{1}{d}. \]

which is the required condition for orthogonal intersection of the given curves.

Two curves touching each other

Prove that the curves

\[ y = x^3 \]

and

\[ y = e^{3(x-1)} \]

touch each other at the point \( (1,1) \).

Solution:

For the curves to touch each other at \( (1,1) \), they must satisfy two conditions:

• The point \( (1,1) \) must satisfy both equations.
• The slopes of their tangents at \( (1,1) \) must be equal.

Substituting \( x = 1 \) in the first curve,

\[ y = 1^3 = 1. \]

For the second curve,

\[ y = e^{3(1-1)} = e^0 = 1. \]

Thus, both curves pass through \( (1,1) \).

Differentiating \( y = x^3 \),

\[ \frac{dy}{dx} = 3x^2. \]

Evaluating at \( x = 1 \),

\[ \left. \frac{dy}{dx} \right|_{x=1} = 3(1)^2 = 3. \]

Differentiating \( y = e^{3(x-1)} \),

\[ \frac{dy}{dx} = e^{3(x-1)} \cdot 3. \]

Evaluating at \( x = 1 \),

\[ \left. \frac{dy}{dx} \right|_{x=1} = e^{3(1-1)} \cdot 3 = e^0 \cdot 3 = 3. \]

Since both curves have the same slope at \( (1,1) \), they have a common tangent at this point. Hence, the curves touch each other at \( (1,1) \).

Shortest Distance Between a Point and a Curve

In many problems, we are interested in finding the shortest distance between a given point \( P(x_0, y_0) \) and a curve defined implicitly by

\[ f(x, y) = 0. \]

The shortest distance always lies along the normal from the given point to the curve. However, in some cases, there may be more than one normal from the given point to the curve, meaning multiple points on the curve can act as the foot of the normal.

To solve this problem, assume the foot of the normal is \( (x_1, y_1) \). This point must satisfy two conditions:

1. Since \( (x_1, y_1) \) lies on the curve, it satisfies

   \[ f(x_1, y_1) = 0. \]

2. The normal at \( (x_1, y_1) \) passes through \( (x_0, y_0) \) and is perpendicular to the tangent at \( (x_1, y_1) \).

   Differentiating \( f(x, y) = 0 \) implicitly, the slope of the tangent at \( (x_1, y_1) \) is

   \[ m_{\text{tangent}} = \frac{dy}{dx} \Big|_{(x_1, y_1)}. \]

   Since the normal is perpendicular to the tangent, its slope satisfies

   \[ m_{\text{tangent}} \cdot \frac{y_1 - y_0}{x_1 - x_0} = -1. \]

This provides two equations:

1. Curve Equation: \( f(x_1, y_1) = 0 \).
2. Normal Condition: \( \frac{dy}{dx} \Big|_{(x_1, y_1)} \cdot \frac{y_1 - y_0}{x_1 - x_0} = -1 \).

Solving these two equations for \( x_1 \) and \( y_1 \), we obtain the possible feet of the normal. Algebraic manipulation may give more than one solution, indicating multiple normal lines from the given point to the curve.

To find the shortest distance, compute the Euclidean distance between \( P(x_0, y_0) \) and each foot of the normal:

\[ d = \sqrt{(x_1 - x_0)^2 + (y_1 - y_0)^2}. \]

Among all possible distances, the smallest one gives the shortest distance between the point and the curve.

Example

A helicopter is flying along the curve

\[ y - x^{3/2} = 7, \quad x \geq 0. \]

A soldier positioned at \( \left(\frac{1}{2}, 7 \right) \) wants to shoot down the helicopter when it is nearest to him. Find this nearest distance.

Solution:

Let the foot of the normal be \( (x_1, y_1) \), where the normal to the curve at this point passes through the soldier's position \( \left(\frac{1}{2}, 7 \right) \). The goal is to determine \( (x_1, y_1) \) and compute the distance between this point and \( \left(\frac{1}{2}, 7 \right) \).

Since \( (x_1, y_1) \) lies on the given curve,

\[ y_1 = x_1^{3/2} + 7. \quad \text{(1)} \]

Differentiating both sides of the equation \( y = x^{3/2} + 7 \),

\[ \frac{dy}{dx} = \frac{3}{2} x^{1/2}. \]

Thus, the slope of the tangent at \( (x_1, y_1) \) is

\[ m_{\text{tangent}} = \frac{3}{2} x_1^{1/2}. \quad \text{(2)} \]

Since the normal is perpendicular to the tangent, the slope of the normal is

\[ m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}} = -\frac{2}{3} x_1^{-1/2}. \quad \text{(3)} \]

Since the normal passes through both \( (x_1, y_1) \) and \( \left(\frac{1}{2}, 7 \right) \), its slope is also given by

\[ m_{\text{normal}} = \frac{y_1 - 7}{x_1 - \frac{1}{2}}. \quad \text{(4)} \]

Equating (3) and (4),

\[ -\frac{2}{3} x_1^{-1/2} = \frac{y_1 - 7}{x_1 - \frac{1}{2}}. \]

Substituting \( y_1 = x_1^{3/2} + 7 \) from (1),

\[ -\frac{2}{3} x_1^{-1/2} = \frac{x_1^{3/2} + 7 - 7}{x_1 - \frac{1}{2}}. \]

\[ -\frac{2}{3} x_1^{-1/2} = \frac{x_1^{3/2}}{x_1 - \frac{1}{2}}. \]

Multiplying both sides by \( 3(x_1 - \frac{1}{2}) \),

\[ -2 x_1^{-1/2} (x_1 - \frac{1}{2}) = 3x_1^{3/2}. \]

Multiplying both sides by \( x_1^{1/2} \),

\[ -2(x_1 - \frac{1}{2}) = 3x_1^2. \]

Rearranging,

\[ 3x_1^2 + 2x_1 - 1 = 0. \]

Factoring,

\[ (3x_1 - 1)(x_1 + 1) = 0. \]

Since \( x_1 \geq 0 \), we take the positive root:

\[ x_1 = \frac{1}{3}. \]

Using (1),

\[ y_1 = \left(\frac{1}{3} \right)^{3/2} + 7. \]

\[ y_1 = \frac{1}{3\sqrt{3}} + 7. \]

Thus, the foot of the normal is

\[ \left(\frac{1}{3}, \frac{1}{3\sqrt{3}} + 7\right). \]

The shortest distance is the Euclidean distance between \( (x_1, y_1) \) and \( \left(\frac{1}{2}, 7 \right) \),

\[ d = \sqrt{\left(\frac{1}{2} - \frac{1}{3} \right)^2 + \left(7 - \left(\frac{1}{3\sqrt{3}} + 7\right) \right)^2 }. \]

\[ = \sqrt{\left(\frac{3 - 2}{6} \right)^2 + \left(\frac{-1}{3\sqrt{3}} \right)^2 }. \]

\[ = \sqrt{\left(\frac{1}{6} \right)^2 + \left(\frac{1}{27} \right)}. \]

\[ = \sqrt{\frac{1}{36} + \frac{1}{27}}. \]

\[ = \frac{\sqrt{7}}{\sqrt{108}} = \frac{\sqrt{7}}{6\sqrt{3}}. \]

\[ = \frac{1}{6} \sqrt{\frac{7}{3}}. \]

Thus, the corrected shortest distance is

\[ \frac{1}{6} \sqrt{\frac{7}{3}}. \]

Shortest Distance Between a Line and a Curve

Consider a given line

\[ L: ax + by + c = 0. \]

Suppose this line does not intersect the curve

\[ f(x, y) = 0. \]

The goal is to determine the shortest distance from the line to the curve. This is defined as the minimum distance between any point \( P \) on the line and any point \( Q \) on the curve, i.e.,

\[ \min \limits_{} d(P, Q). \]

This minimum distance occurs when PQ is normal to both the line and the curve. This is equivalent to saying the tangent at Q is parallel to the given line L. The approach is to first find the foot of the normal \( Q(x_1, y_1) \) on the curve from the line.

Since there are two unknowns \( x_1 \) and \( y_1 \), we need two independent equations to determine their values:

1. Curve Condition: Since \( Q(x_1, y_1) \) lies on the curve, it must satisfy the equation

   \[ f(x_1, y_1) = 0. \quad \text{(1)} \]

2. Parallel Tangent Condition:

   • Differentiating the curve implicitly, the slope of the tangent at \( Q(x_1, y_1) \) is

     \[ m_{\text{tangent}} = \frac{dy}{dx} \Big|_{(x_1, y_1)}. \]

   • The slope of the given line is \( -\frac{a}{b} \).

   • Since the tangent at \( Q(x_1, y_1) \) is parallel to the given line, their slopes must be equal:

     \[ \frac{dy}{dx} \Big|_{(x_1, y_1)} = -\frac{a}{b}. \quad \text{(2)} \]

Solving (1) and (2) simultaneously gives the coordinates \( (x_1, y_1) \).

Once \( (x_1, y_1) \) is found, the shortest distance is simply the perpendicular distance from \( Q(x_1, y_1) \) to the given line, given by

\[ d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}. \]

Thus, the shortest distance between the line and the curve is found by determining the foot of the normal and computing its perpendicular distance to the line.

Example

Find the shortest distance of the line

\[ x - 4y + 5 = 0 \]

from the curve

\[ y = \sqrt{x - 1}, \quad x \geq 1. \]

Solution:

Let PQ be a common normal to the given line and the curve, where \( P \) lies on the line and \( Q(x_1, y_1) \) lies on the curve. Since \( Q \) lies on the curve, it must satisfy

\[ y_1 = \sqrt{x_1 - 1}. \quad \text{(1)} \]

Differentiating the curve implicitly,

\[ \frac{dy}{dx} = \frac{1}{2\sqrt{x - 1}}. \]

Thus, at \( Q(x_1, y_1) \),

\[ \frac{dy}{dx} = \frac{1}{2\sqrt{x_1 - 1}}. \quad \text{(2)} \]

The slope of the given line is obtained by rewriting its equation in slope-intercept form:

\[ y = \frac{1}{4}x + \frac{5}{4}. \]

Thus, its slope is

\[ m = \frac{1}{4}. \]

Since the shortest distance occurs when the tangent at \( Q(x_1, y_1) \) is parallel to the given line, equating slopes from (2),

\[ \frac{1}{2\sqrt{x_1 - 1}} = \frac{1}{4}. \]

Solving for \( x_1 \),

\[ 2\sqrt{x_1 - 1} = 4. \]

\[ \sqrt{x_1 - 1} = 2. \]

\[ x_1 - 1 = 4. \]

\[ x_1 = 5. \]

Substituting \( x_1 = 5 \) in equation (1),

\[ y_1 = \sqrt{5 - 1} = \sqrt{4} = 2. \]

Thus, the foot of the normal is \( Q(5,2) \).

The shortest distance is the perpendicular distance from \( Q(5,2) \) to the given line \( x - 4y + 5 = 0 \), given by

\[ d = \frac{|(1)(5) + (-4)(2) + 5|}{\sqrt{1^2 + (-4)^2}}. \]

\[ = \frac{|5 - 8 + 5|}{\sqrt{1 + 16}}. \]

\[ = \frac{|2|}{\sqrt{17}}. \]

\[ = \frac{2}{\sqrt{17}}. \]

Thus, the required shortest distance is

\[ \frac{2}{\sqrt{17}}. \]

Shortest Distance Between a Circle and a Curve

The shortest distance between a circle and a curve refers to the minimum length of a segment PQ, where \( P \) lies on the circle and \( Q \) lies on the curve. The key observation is that PQ is minimized when it is a common normal to both the circle and the curve.

A circle has a special property: the normal at any point on the circle always passes through the center of the circle. Let the center of the circle be \( C(h, k) \), and let the radius of the circle be \( R \).

To determine the shortest distance \( PQ \), observe that it is equivalent to computing \( CQ \) first, where Q is the foot of the normal from \( C \) to the curve.

Thus, the problem reduces to finding the shortest distance from the center \( C(h, k) \) to the curve, which has already been discussed previously.

1. Find \( Q(x_1, y_1) \), the foot of the normal from \( C(h, k) \) to the curve.

   • The point \( Q(x_1, y_1) \) satisfies the equation of the curve:

     \[ f(x_1, y_1) = 0. \]

   • The slope of the normal at \( Q(x_1, y_1) \) is given by

     \[ \frac{dy}{dx} \Big|_{(x_1, y_1)}. \]

   • Since \( CQ \) is the normal, it must be perpendicular to the tangent at \( Q \). The perpendicularity condition gives the second equation:

     \[ \frac{dy}{dx} \Big|_{(x_1, y_1)} \cdot \left(\frac{y_1 - k}{x_1 - h}\right) =-1. \]

2. Solve these equations for \( x_1, y_1 \).

3. Compute \( CQ \), the shortest distance from \( C(h, k) \) to the curve, using the Euclidean formula:

   \[ CQ = \sqrt{(x_1 - h)^2 + (y_1 - k)^2}. \]

4. Compute the required shortest distance between the circle and the curve:

   \[ PQ = CQ - R. \]

Thus, the shortest distance between a circle and a curve is the shortest distance from the center of the circle to the curve, minus the radius of the circle.

Shortest Distance Between Two General Curves

The shortest distance between two given curves \( f(x, y) = 0 \) and \( g(x, y) = 0 \) is defined as the minimum possible distance between a point \( P(x_1, y_1) \) lying on the first curve and a point \( Q(x_2, y_2) \) lying on the second curve. That is,

\[ \min \limits_{(x_1, y_1) \in f(x, y) = 0, \, (x_2, y_2) \in g(x, y) = 0} d(P, Q). \]

Key Observation: Common Normal or Parallel Tangents

The shortest distance occurs when PQ is normal to both curves. This means that the tangent at \( P(x_1, y_1) \) is parallel to the tangent at \( Q(x_2, y_2) \).

Procedure to Find the Shortest Distance

Since there are four unknowns, \( x_1, y_1, x_2, y_2 \), we require four independent equations to determine them. These equations are obtained as follows:

1. Curve Condition at \( P(x_1, y_1) \): Since \( P(x_1, y_1) \) lies on the first curve, it must satisfy

   \[ f(x_1, y_1) = 0. \quad \text{(1)} \]

2. Curve Condition at \( Q(x_2, y_2) \): Since \( Q(x_2, y_2) \) lies on the second curve, it must satisfy

   \[ g(x_2, y_2) = 0. \quad \text{(2)} \]

3. Slope Condition for Parallel Tangents: The tangents at \( P \) and \( Q \) must be parallel. Differentiating \( f(x, y) = 0 \) and \( g(x, y) = 0 \) implicitly, the slopes of their tangents at \( P \) and \( Q \) are

   \[ \frac{dy}{dx} \Big|_{(x_1, y_1)} = \frac{dy}{dx} \Big|_{(x_2, y_2)}. \quad \text{(3)} \]

4. Normal Condition: The shortest distance line \( PQ \) is perpendicular to the common tangent, meaning its slope is the negative reciprocal of the common tangent slope. That is,

   \[ \frac{y_2 - y_1}{x_2 - x_1} = -\frac{1}{\frac{dy}{dx} \Big|_{(x_1, y_1)}}. \quad \text{(4)} \]

Solving (1), (2), (3), and (4) simultaneously gives the points \( (x_1, y_1) \) and \( (x_2, y_2) \), which define the shortest distance.

Complexity of the General Case

In most general cases, solving these four nonlinear equations algebraically is extremely difficult. Therefore, additional information is often provided or used to simplify the problem.

Alternatively, if one of the curves is a straight line or a circle, then the problem reduces to previously discussed special cases, which are computationally simpler.

Example

Find the shortest distance between the curves

\[ y = e^x \quad \text{and} \quad y = \ln x. \]

Solution:

Observe that the curve \( y = e^x \) is the reflection of the curve \( y = \ln x \) in the line \( y = x \). Due to this symmetry, the shortest distance between the two curves is simply twice the shortest distance between the line \( y = x \) and the curve \( y = e^x \).

To find this shortest distance, let the foot of the common normal between the line \( y = x \) and the curve \( y = e^x \) be \( N(x_1, y_1) \). Since \( N \) lies on the curve, it satisfies

\[ y_1 = e^{x_1}. \quad \text{(1)} \]

The slope of the tangent at \( N \) is

\[ \frac{dy}{dx} = e^{x_1}. \]

For the shortest distance, the tangent to the curve must be parallel to the line \( y = x \), which has a slope of 1. Equating the slopes,

\[ e^{x_1} = 1. \]

Solving for \( x_1 \),

\[ x_1 = 0. \]

Substituting in (1),

\[ y_1 = e^0 = 1. \]

Thus, the foot of the normal is \( N(0,1) \). The shortest distance between \( N(0,1) \) and the line \( y = x \) is the perpendicular distance:

\[ d = \frac{|0 - 1|}{\sqrt{1^2 + (-1)^2}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}. \]

Since the shortest distance between \( y = e^x \) and \( y = \ln x \) is twice this distance, the final answer is

\[ 2 \times \frac{\sqrt{2}}{2} = \sqrt{2}. \]