Question

The locus of the point of intersection of tangents to the parabola $y^2=4(x+1)$ and $y^2=8(x+2)$which are perpendicular to each other is

• A. $x+7=0$
• B. $x-y=A$
• C. $x+3=0$
• D. $y-x=12.$

Answer 1

Answer: (C)

Any tangent to $y^2=4\left(x+1\right)$ is
$y=m\left(x+1\right)+\frac{1}{m}...\left(1\right)$
Any tangent to $y^2=8\left(x+2\right)$is
$y=m^{\prime }\left(x+2\right)+\frac{2}{m}...\left(2\right)$
Since $\left(1\right)$ and $\left(2\right)$ are $\perp$
$\therefore m{m}^{\prime }=-1$
$\therefore m=-\frac{1}{m}$
Putting in $\left(2\right)$, we get
$y=-\frac{1}{m}\left(x+2\right)-2m...\left(3\right)$
$\left(1\right)-\left(3\right)$, gives
$0=\left(m+\frac{1}{m}\right)x+m+\frac{1}{m}+\frac{2}{m}+2m$
i.e., $0=\left(m+\frac{1}{m}\right)x+3\left(m+\frac{1}{m}\right)$
i.e., $0=\left(x+3\right)\left(m+\frac{1}{m}\right)$
$\Rightarrow x+3=0\left(\because m+\frac{1}{m}\ne 0\right)$