Question

The locus of the point of intersection of two normals to the parabola $x ^ { 2 } = 8 y$ which are at right angles to each other, is

• A. $x ^ { 2 } = 2 ( y - 6 )$
• B. $x ^ { 2 } = 2 ( y + 6 )$
• C. $x ^ { 2 } = - 2 ( y - 6 )$
• D. None of these

Answer 1

Answer: (A)

$x ^ { 2 } = 2 ( y - 6 )$

Answer 2

Given parabola is $x ^ { 2 } = 8 y$ .....(i)

Let $Q \left( 4 t _ { 2 } , 2 t _ { 2 } ^ { 2 } \right)$be two points on the parabola (i)

Normal at P, Q are $y - 2 t _ { 1 } ^ { 2 } = - \frac { 1 } { t _ { 1 } } \left( x - 4 t _ { 1 } \right)$ ......(ii) and

$y - 2 t _ { 2 } ^ { 2 } = - \frac { 1 } { t _ { 2 } } \left( x - 4 t _ { 2 } \right)$ ......(iii)

(ii)–(iii) gives $2 \left( t _ { 2 } ^ { 2 } - t _ { 1 } ^ { 2 } \right) = x \left( \frac { 1 } { t _ { 2 } } - \frac { 1 } { t _ { 1 } } \right)$ $= x \frac { t _ { 1 } - t _ { 2 } } { t _ { 1 } t _ { 2 } }$,

∴$x = - 2 t _ { 1 } t _ { 2 } \left( t _ { 2 } + t _ { 1 } \right)$ . ....(iv)

From (ii), $y = 2 t _ { 1 } ^ { 2 } - \frac { 1 } { t _ { 1 } } \left( - 2 t _ { 1 } t _ { 2 } \left( t _ { 2 } + t _ { 1 } \right) - 4 t _ { 1 } \right)$

$y = 2 t _ { 1 } ^ { 2 } + 2 t _ { 1 } t _ { 2 } + 2 t _ { 2 } ^ { 2 } + 4$ .....(v)

Since normals (ii) and (iii) are at right angles, ∴ $t _ { 1 } t _ { 2 } = - 1$

∴ From (iv), $x = 2 \left( t _ { 1 } + t _ { 2 } \right)$ and from (v) $= 2 \left[ t _ { 1 } ^ { 2 } + t _ { 2 } ^ { 2 } + 1 \right]$ $= 2 \left[ \left( t _ { 1 } + t _ { 2 } \right) ^ { 2 } - 2 t _ { 1 } t _ { 2 } + 1 \right]$

⇒ y$= 2 \left[ \left( t _ { 1 } + t _ { 2 } \right) ^ { 2 } + 3 \right.$

⇒$y = 2 \left[ \frac { x ^ { 2 } } { 4 } + 3 \right] = \frac { x ^ { 2 } } { 2 } + 6$⇒ $x ^ { 2 } = 2 ( y - 6 )$ which is the

required locus.