Search any question & find its solution
Question:
Answered & Verified by Expert
If the line segment joining the vertex of the parabola $y^2=4 a x$ and a point on the parabola, makes an angle $\theta$ with the positive $X$-axis, then the length of that line segment is
Options:
Solution:
1415 Upvotes
Verified Answer
The correct answer is:
$\frac{4 a \cos \theta}{\sin ^2 \theta}$
Equation of given parabola is $y^2=4 a x$ having
vertex $V(0,0)$
Let a point on the parabola $P\left(a t^2, 2 a t\right)$, so slope of line joining point $V(0,0)$ and $P\left(a t^2, 2 a t\right)$ is
$\frac{2 a t-0}{a t^2-0}=\tan \theta$
(given)
$\Rightarrow \quad \frac{2}{t}=\tan \theta \Rightarrow t=2 \cot \theta$
Now, length of line segment $V P=\sqrt{\left(a t^2\right)^2+(2 a t)^2}$
$\begin{aligned} & =a \sqrt{t^4+4 t^2}=a \sqrt{(2 \cot \theta)^4+4(2 \cot \theta)^2} \\ & =4 a \cot \theta \operatorname{cosec} \theta=\frac{4 a \cos \theta}{\sin ^2 \theta}\end{aligned}$
Hence, option (b) is correct.
vertex $V(0,0)$
Let a point on the parabola $P\left(a t^2, 2 a t\right)$, so slope of line joining point $V(0,0)$ and $P\left(a t^2, 2 a t\right)$ is
$\frac{2 a t-0}{a t^2-0}=\tan \theta$
(given)
$\Rightarrow \quad \frac{2}{t}=\tan \theta \Rightarrow t=2 \cot \theta$
Now, length of line segment $V P=\sqrt{\left(a t^2\right)^2+(2 a t)^2}$
$\begin{aligned} & =a \sqrt{t^4+4 t^2}=a \sqrt{(2 \cot \theta)^4+4(2 \cot \theta)^2} \\ & =4 a \cot \theta \operatorname{cosec} \theta=\frac{4 a \cos \theta}{\sin ^2 \theta}\end{aligned}$
Hence, option (b) is correct.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.