Question

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

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

Answer 1

Answer: (C)

$x+3=0$

The equation of any tangent to $y^2=4(x+1)$ is $y=m(x+1)+\frac{1}{m}$ ...(1)
The equation of any tangent to $y^2=8(x+2)$ is $y=m^{\prime }(x+2)+\frac{2}{m^{\prime }}$ ...(2)
It is given that (1) and (2) are perpendicular. Therefore,
$m{m}^{\prime }=-1\Rightarrow m^{\prime }=-\frac{1}{m}$
Putting $m^{\prime }=-\frac{1}{m}$ in (2), we get
$y=(-\frac{1}{m})(x+2)-2m$ ...(3)
The point of intersection of (1) and (3) is given by solving (1) and (2)
On subtracting (3) from (1), we get
$0=(m+\frac{1}{m})x+3(m+\frac{1}{m})\Rightarrow x+3=0$ $[\because m+\frac{1}{m}\ne 0]$