Equation of Normal to a Parabola
Equation of Normal to a Standard Parabola at a Given Point
Let \(P(x_1, y_1)\) be a point lying on the parabola whose standard equation is \(y^2 = 4ax\). We wish to find the equation of the normal line to the curve at this specific point.
From our study of tangents, the slope of the tangent at \(P(x_1, y_1)\) is:
Since the normal line is perpendicular to the tangent at the point of contact, the slope of the normal at \(P(x_1, y_1)\) is the negative reciprocal of the tangent's slope:
Using the point-slope form of a straight line, \(y - y_1 = m(x - x_1)\), the equation of the normal at \(P(x_1, y_1)\) is:
Similarly, for all four standard orientations of the parabola, the equation of the normal at \((x_1, y_1)\) is summarized below:
| Equation of Parabola | Point of Contact | Equation of Normal |
|---|---|---|
| \(y^2 = 4ax\) | \((x_1, y_1)\) | \(y - y_1 = -\frac{y_1}{2a}(x - x_1)\) |
| \(y^2 = -4ax\) | \((x_1, y_1)\) | \(y - y_1 = \frac{y_1}{2a}(x - x_1)\) |
| \(x^2 = 4ay\) | \((x_1, y_1)\) | \(x - x_1 = -\frac{x_1}{2a}(y - y_1)\) |
| \(x^2 = -4ay\) | \((x_1, y_1)\) | \(x - x_1 = \frac{x_1}{2a}(y - y_1)\) |
Equation of Normal in Parametric Form
We know that any point on the standard parabola \(y^2 = 4ax\) can be represented in parametric form as \((at^2, 2at)\), where \(t\) is a parameter. We will now derive the equation of the normal line at this parametric point.
To find the equation of the normal, we first determine its slope. Since the normal line is perpendicular to the tangent line at the point of contact, and the slope of the tangent to the parabola at \((at^2, 2at)\) is:
the slope of the normal (\(m_{\text{normal}}\)) is the negative reciprocal of the tangent's slope:
Now, using the slope \(m = -t\) and the point \((at^2, 2at)\) in the point-slope form \(y - y_1 = m(x - x_1)\), we obtain:
Distributing \(-t\) on the right-hand side yields:
Rearranging the terms, we arrive at the standard parametric equation of the normal to the parabola:
This equation is remarkably useful because it defines the normal entirely in terms of the single parameter \(t\).
| Equation of Parabola | Parametric Point | Equation of Normal |
|---|---|---|
| \(y^2 = 4ax\) | \((at^2, 2at)\) | \(y + tx = 2at + at^3\) |
| \(y^2 = -4ax\) | \((-at^2, 2at)\) | \(y - tx = 2at + at^3\) |
| \(x^2 = 4ay\) | \((2at, at^2)\) | \(x + ty = 2at + at^3\) |
| \(x^2 = -4ay\) | \((2at, -at^2)\) | \(x - ty = 2at + at^3\) |
Condition for a Line to be Normal to a Parabola (Slope Form)
Consider a general line \(y = mx + c\). For this line to be normal to the parabola \(y^2 = 4ax\), it must coincide with the parametric normal for some parameter \(t\).
We know that the equation of the normal in parametric form is:
Comparing this with the line \(y = mx + c\), we identify:
and
Substituting \(t = -m\) into the expression for \(c\), we obtain the condition of normality:
Equation of Normal in Slope Form
The equation of the normal to the parabola \(y^2 = 4ax\) having a given slope \(m\) is:
The corresponding point of contact (the foot of the normal) is \((at^2, 2at) = (am^2, -2am)\).
Slope Form for Vertical Parabolas
For the upward-opening vertical parabola \(x^2 = 4ay\), the equation of the normal in slope form is:
and the foot of the normal is \((2at, at^2) = \left(-\frac{2a}{m}, \frac{a}{m^2}\right)\), where \(m = -\frac{1}{t}\).
Let us derive this directly using calculus and coordinate geometry so that the method is completely clear.
Let the equation of the normal be \(y = mx + c\), and let \((x_1, y_1)\) be the foot of the normal on the parabola \(x^2 = 4ay\).
Differentiating \(x^2 = 4ay\) with respect to \(x\):
The slope of the tangent at \((x_1, y_1)\) is \(\frac{x_1}{2a}\). Therefore, the slope \(m\) of the normal is the negative reciprocal:
Since the point \((x_1, y_1)\) lies on the parabola \(x^2 = 4ay\):
Thus, the foot of the normal is \(\left(-\frac{2a}{m}, \frac{a}{m^2}\right)\).
Now, substituting the coordinates of this point into the line equation \(y = mx + c\):
Substituting \(c\) back into \(y = mx + c\), we obtain the equation of the normal in slope form:
| Equation of Parabola | Foot of the Normal | Equation of Normal (Slope Form) |
|---|---|---|
| \(y^2 = 4ax\) | \((am^2, -2am)\) | \(y = mx - 2am - am^3\) |
| \(y^2 = -4ax\) | \((-am^2, 2am)\) | \(y = mx + 2am + am^3\) |
| \(x^2 = 4ay\) | \(\left(-\frac{2a}{m}, \frac{a}{m^2}\right)\) | \(y = mx + 2a + \frac{a}{m^2}\) |
| \(x^2 = -4ay\) | \(\left(\frac{2a}{m}, -\frac{a}{m^2}\right)\) | \(y = mx - 2a - \frac{a}{m^2}\) |
Solved Examples
Example 1
If the normal at the point \(P(at_1^2, 2at_1)\) on the parabola \(y^2 = 4ax\) meets the curve again at the point \(Q(at_2^2, 2at_2)\), prove that:
Hence, show that if the normal chord \(PQ\) subtends a right angle at the vertex of the parabola, then \(t_1 = \pm\sqrt{2}\) and the normal is inclined at an angle \(\tan^{-1}\sqrt{2}\) to the axis of the parabola.
Solution
Part 1: Relation between \(t_1\) and \(t_2\)
The equation of the normal to the parabola \(y^2 = 4ax\) at the point \(P(at_1^2, 2at_1)\) is:
Since this normal meets the parabola again at the point \(Q(at_2^2, 2at_2)\), the coordinates of \(Q\) must satisfy the equation of the normal:
Grouping similar terms on each side gives:
Factoring \(t_2^2 - t_1^2 = (t_2 - t_1)(t_2 + t_1)\):
Since \(P\) and \(Q\) are distinct points, \(t_2 \neq t_1\), and since \(a \neq 0\), we can divide both sides by \(a(t_2 - t_1)\):
Part 2: Normal chord subtending a right angle at the vertex
The vertex of the parabola \(y^2 = 4ax\) is the origin \(O(0, 0)\).
The slope of the chord \(OP\) joining \((0,0)\) and \(P(at_1^2, 2at_1)\) is:
Similarly, the slope of the chord \(OQ\) joining \((0,0)\) and \(Q(at_2^2, 2at_2)\) is:
Since the normal chord \(PQ\) subtends a right angle at the vertex \(O\), the lines \(OP\) and \(OQ\) are perpendicular (\(m_{OP} \cdot m_{OQ} = -1\)):
Substituting \(t_2 = -\frac{4}{t_1}\) into our normal chord relation \(t_2 = -t_1 - \frac{2}{t_1}\):
Finally, since the slope of the normal at \(P(t_1)\) is \(m = -t_1 = \mp\sqrt{2}\), the acute angle \(\theta\) that the normal makes with the axis of the parabola (\(x\)-axis) satisfies \(\tan\theta = \sqrt{2}\), so \(\mathbf{\theta = \tan^{-1}\sqrt{2}}\).
Example 2
Prove that the locus of the point of intersection of two mutually perpendicular normals to the parabola \(y^2 = 4ax\) is the parabola:
Solution
The equation of any normal to the parabola \(y^2 = 4ax\) in slope form is:
Let \((h, k)\) be the point of intersection of two mutually perpendicular normals. Since the normal passes through \((h, k)\), its coordinates satisfy the equation:
Rearranging this into a cubic equation in \(m\):
For a given point \((h, k)\), the three roots \(m_1, m_2, m_3\) of this cubic equation represent the slopes of the three normals that can be drawn from \((h, k)\) to the parabola. By Vieta's formulas, the sum and product of the roots are:
We are given that two of the normals passing through \((h, k)\) are mutually perpendicular. Let their slopes be \(m_1\) and \(m_2\), so that:
Substituting \(m_1 m_2 = -1\) into equation (3) immediately gives the third root \(m_3\):
Since \(m_3 = \frac{k}{a}\) is a root of the cubic equation (1), it must satisfy equation (1):
Assuming \(k \neq 0\), we divide the entire equation by \(\frac{k}{a}\):
Replacing \((h, k)\) with the general coordinates \((x, y)\), we obtain the required locus: