Question

The length of the line segment joining the vertex of the parabola $y^2=4ax$ and a point on the parabola where the line segment makes an angle $\theta$ to the $x$ -axis is $\frac{4am}{n}.$ Here, $m$ and $n$ respectively are

• A. $\sin \theta ,\cos \theta$
• B. $\cos \theta ,\sin \theta$
• C. $\cos \theta ,{\sin }^2\theta$
• D. ${\sin }^2\theta ,\cos \theta$

Answer 1

Answer: (C)

Let any point $P(h,k)$ will satisfy $y^2=4ax$ i.e, $k^2=4ah$
Let a line $OP$ makes an angle $\theta$ from the $x$ -axis.
$\therefore \text{ In }\Delta OAP,\sin \theta =\frac{PA}{OP}$
$\sin \theta =\frac{k}{l}\Rightarrow k=l\sin \theta$
and $\cos \theta =\frac{OA}{OP}\Rightarrow \cos \theta =\frac{h}{l}\Rightarrow h=\ell \cos \theta$

Hence, from eq.(i), we get
$l^2{\sin }^2\theta =4a\times l\cos \theta ($ put $k=l\sin \theta ,h=l\cos \theta )$
$\Rightarrow l=\frac{4a\cos \theta }{{\sin }^2\theta }$