If a normal chord at a point t on the parabola y2=4 a x subtends a right angle at the vertex, then t equals to

Question

If a normal chord at a point t on the parabola y2=4 a x subtends a right angle at the vertex, then t equals to (A) 1 (B) √2 (C) 2 (D) √3

Tardigrade Admin · Accepted Answer

The perpendicular of the normal to the parabola

y^{2} = 4 a x

at

P

is

Suppose, it meets the parabola at

Q

. If

O

be the vertex of the parabola, then the combined equation of

OP

and

OQ

is a homogeneous equation of second degree.

y^{2} = 4 a x (\frac{y + t x}{2 a t + a t ^{3}})

\Rightarrow y^{2} (2 a t + a t^{3}) = 4 a x (y + t x)

\Rightarrow 4 a t x^{2} + 4 a x y - (2 a t + a t^{3}) y^{2} = 0

Since,

OP

and

OQ

are at right angles,
then Coefficient of

x^{2} +

Coefficient of

y^{2} = 0

∴ 4 a t - 2 a t - a t^{3} = 0

\Rightarrow t^{2} = 2

\Rightarrow t = 2

Q. If a normal chord at a point $t$ on the parabola $y^{2} = 4 a x$ subtends a right angle at the vertex, then $t$ equals to