Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

If a normal at $\mathrm{P\left(a t^2, 2 a t\right)}$ to $\mathrm{ y^2=4 a x}$ intersects the curve at Q and if the line segment PQ subtends a right angle at the vertex, then
Option: 1
$\mathrm{t^2=2}$

Option: 2
$\mathrm{t^2=1}$

Option: 3
$\mathrm{t=1}$

Option: 4
$\mathrm{t^2+1=0}$

Answers (1)

best_answer

The normal at $\mathrm{P\left(a t^2, 2 a t\right)}$ is

$\mathrm{ \begin{aligned} & y-2 a t=-t\left(x-a t^2\right) \\ & y+t x=2 a t+a t^3 \end{aligned} }$ $\mathrm{..................(1)}$

Let it cut the parabola at Q.

As PQ subtends $\mathrm{90^{\circ}}$ at the vertex, by homogeneity method, the combined equation of OP, OQ is

$\mathrm{ y^2-4 a x\left(\frac{y+t x}{2 a t+a t^3}\right)=0 }$

Or $\mathrm{y^2\left(2 a t+a t^3\right)=4 a x(y+t x)}$ $\mathrm{..................(2)}$

As $\mathrm{O P \perp O Q,}$ the condition is that

the sum of coefficients of $\mathrm{x^2 }$ and $\mathrm{y^2}$ in (2) is zero

Or $\mathrm{\left(2 a t+a t^3\right)-4 a t=0}$

Or $\mathrm{a t^3-2 a t=0}$

As $\mathrm{\quad t \neq 0 \Rightarrow t^2=2}$

Posted by

Sayak

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Latest Question