Problem :
If the segment intercepted by the parabola $y^2 =4ax$ with the line lx +my +n=0 subtends a right angle at the vertex, then
(a) 4al +n=0
(b)4am +n=0
(c) al +n=0
(d) 4al +4am +n=0
My working :
Let the point of intersection and the parabola is A (a,b) and B (c,d) therefore it should satisy the equation of line : $\Rightarrow al +mb +n=0 …….(i) ; cl +md +n=0 ……(ii)$
Also the vertex of the parabola is at the origin O(0,0) and the point O(0,0) ; A(a,b) and O(0,0) ; B(c,d) are at right perpendicular to each other
$\Rightarrow \frac{b}{a} \times \frac{d}{c} = -1 ……….(iii)$
Also AOB forms a right angle triangle therefore : $AO^2 + BO^2 = AB^2$
Is this the right way of approaching this problem or is there some shorter method to solve this please suggest..thanks..
Best Answer
The parabola $y^2=4ax$ can be written parametrically as $x=at^2,y=2at$
Let $\displaystyle P(au^2,2au), Q(av^2,2av)$ be two intersections and $O(0,0)$ be the vertex with $u\cdot v\ne0$
As $\displaystyle PO\perp OQ,$ $$\frac{2au-0}{au^2-0}\cdot \frac{2av-0}{av^2-0}=-1\implies uv=-4\ \ \ \ (1)$$
Now $P,Q$ lies on the straight line $\displaystyle lx+my+n=0$
So, $\displaystyle l(au^2)+m(2au)+n=0\implies lau^2+2mau+n=0$ and $\displaystyle lav^2+2mav+n=0$
Clearly, $u,v$ are the roots of the Quadratic equation $\displaystyle lat^2+2mat+n=0$
$\displaystyle\implies uv=\frac n{la}\ \ \ \ (2)$
Compare $(1),(2)$
Just to remind the condition derived is only necessary, but not sufficient for perpendicularity/Orthogonality or $PO,OQ$