Question

If the segment intercepted by the parabola $y=4ax$ with the line $lx+my+n=0$ subtends a right angle at the vertex then

• A. $4al+n=0$
• B. $4al+4am+n=0$
• C. $4am+n=0$
• D. $al+n=0$

Answer 1

The correct option is A $4al+n=0$

The parabola $y^2=3ax$ can be written parametrically as $x=a{t}^2,y=at$

Let $P(a{u}^2,2au),Q(a{v}^2,2av)$ be two intersections and $O(0,0)$ be the vertex with $u.v\ne 0$

As $PO\perp OQ$

Product of slopes=-1

$m_1{m}_2=-1$

$\frac{2au-0}{a{u}^2-0}\cdot \frac{2av-0}{a{v}^2-0}=-1$

$uv=-\mathrm{4.....}\left(1\right)$

Now $P,Q$ lies on the straight line $lx+my+n=0$

So,

$l\left(a{u}^2\right)+m\left(2au\right)+n=0$

$la{u}^2+2mau+n=0$ and $la{v}^2+2mav+n=0$

Clearly, $u,v$ are the roots of the quadratic equation $la{t}^2+2mav+n=0$$

$uv=\frac{n}{la}.....\left(2\right)$

Compare (1) and (2)

$-4=\frac{n}{la}$

$-4la=n$

$4la+n=0$