First slide

Parabola

Question

Locus of the point of intersection of the normals to the parabola $y^{2} = 16 x$ which are at right angles is

Moderate

A

$y^{2} = 4 (x - 4)$
B

$y^{2} = 4 (x - 8)$
C

$y^{2} = - 4 (x - 12)$
D

$y^{2} = - 8 (x - 8)$

Solution

Let two perpendicular normals to the parabola $y^{2} = 16 x$
be $y = m x - 8 m - 4 m^{3}$ and $y = - \frac{1}{m} x + \frac{8}{m} + \frac{4}{m^{3}}$
Solving we get $x = 4 {(m - \frac{1}{m})}^{2} + 12$
$y = - 4 (m - \frac{1}{m})$
Eliminating $m$ we get the required locus as
$y^{2} = 4 (x - 12)$ .

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