Question

Let $L_1$ be a tangent to the parabola $y^2=4(x+1)$ and $L_2$ be a tangent to the parabola $y^2=8(x+2)$ such that $L_1$ and $L_2$ intersect at right angles. Then $L_1$ and $L_2$ meet on the straight line:-

• A. $x+3=0$
• B. $x+2y=0$
• C. $2x+1=0$
• D. $x+2=0$

Answer 1

Answer: (A)

$y^2=4(x+1)$
equation of tangent $y=m(x+1)+\frac{1}{m}$
$y=mx+m+\frac{1}{m}$
$y^2=8(x+2)$
equation of tangent $y=m^{\prime }(x+2)+\frac{2}{m^{\prime }}$
$y=m^{\prime }x+2\left(m^{\prime }+\frac{1}{m^{\prime }}\right)$
since lines intersect at right angles
$\therefore m{m}^{\prime }=-1$
Now $y=mx+m+\frac{1}{m}....(1)$
$y=m^{\prime }x+2\left(m^{\prime }+\frac{1}{m^{\prime }}\right)$
$y=-\frac{1}{m}x+2\left(-\frac{1}{m}-m\right)$
$y=-\frac{1}{m}x-2\left(m+\frac{1}{m}\right)....(2)$
From equation (1) and (2)
$mx+m+\frac{1}{m}=-\frac{1}{m}x-2\left(m+\frac{1}{m}\right)$
$\left(m+\frac{1}{m}\right)x+3\left(m+\frac{1}{m}\right)=0$
$\therefore x+3=0$