Question

If the line $lx+my+n=0$ touches the parabola $y^2=4a(x-b)$ then

• A. $a{l}^2=b{m}^2+nl$
• B. $a{m}^2=b{l}^2+nl$
• C. $a{n}^2=b{l}^2+nm$
• D. $a{n}^2=b{l}^2+2nm$

Answer 1

The correct option is B $a{m}^2=b{l}^2+nl$

Given eqn of line is
$lx+my+n=0$ .....(1)
Given eqn of parabola
$y^2=4a(x-b)$ .....(2)
Let $P(x_1,y_1)$ be the point where the line touches the parabola
Since, $P$ lies on the line
$\therefore l{x}_1+m{y}_1+n=0$
$\Rightarrow x_1=\frac{-m{y}_1-n}{l}$
Also, $P$ lies on the parabola
$\therefore y_1^2=4a(x_1-b)$
$\Rightarrow y_1^2=4a(\frac{-m{y}_1-n}{l}-b)$
$\Rightarrow l{y}_1^2+4am{y}_1+4a(n+bl)$
As we know that the line touches this parabola if the eqn has equal roots

i.e. discriminant $D=0$ i.e. $b^2-4ac=0$

$\Rightarrow 16a^2{m}^2-4l(4an+4abl)=0$
$\Rightarrow 16a^2{m}^2=4l(4an+4abl)$
$\Rightarrow a{m}^2=ln+b{l}^2$