Question

In the parabola $y^2=4ax$, the tangent at the point $P$, whose abscissa is equal to the latus rectum meets the axis in $T$ and the normal at $P$ cuts the parabola again in $Q$ then $PT:PQ=3:5$.

• A. True
• B. False

Answer 1

The correct option is B False

Equation of the parabola is $y^2=4ax$(given)

Latus rectum$=4a$

Let $P(a{t}^2,2at)$ be the coordinates of tangent.

Given: Abscissa of the tangent at $P=$ latus rectum meeting its axis at $T$

$\Rightarrow a{t}^2=4a$

$\Rightarrow t^2=4$

$\Rightarrow t=\pm 2$

Take $t=2$ then coordinates of tangent are $P(4a,4a)$

Tangent at the point $P$ is $2y=x+4a$ meets the axis in $T$

$\Rightarrow y=0$

$\Rightarrow x+4a=0$

$\Rightarrow x=-4a$

$\therefore T$ is $(-4a,0)$

Normal at $P$ cuts at $Q(a{t_1}^2,2a{t}_1)$

where $t_1=-t-\frac{-2}{t}=-2-\frac{-2}{2}=-2-1=-3$

$\therefore Q$ is $(9a,-6a)$

Now,${\left(PQ\right)}^2={(-6a-0)}^2+{(9a-4a)}^2=36a^2+169a^2=125a^2$

${\left(PT\right)}^2={(4a+4a)}^2+{(4a-0)}^2=64a^2+16a^2=80a^2$

Ratio$=\frac{{\left(PT\right)}^2}{{\left(PQ\right)}^2}=\frac{80a^2}{125a^2}=\frac{16}{25}$

$\Rightarrow \frac{PT}{PQ}=\frac{4}{5}$

$\therefore PT:PQ=4:5$

Hence the given statement is false.