Question

The locus of the point of intersection of perpendicular tangent drawn to each one of the parabola $y^2=4x+4$ and $y^2=8x+16$ is

• A. $x=-3$
• B. $x=-12$
• C. $x=-8$
• D. $x=-4$

Answer 1

The correct option is A $x=-3$

Given parabolas are $y^2=4x+4$ and $y^2=8x+16$
Let the slope of one tangent be $m$, so the slope of other tangent is $-\frac{1}{m}$
$(\because$ They are perpendicular to each other$)$
Now, the equation of tangent to the parabola $y^2=4x+4$ is
$y=m(x+1)+\frac{1}{m}\cdots \left(1\right)$
The equation of tangent to parabola $y^2=8x+16$ is
$y=-\frac{1}{m}(x+2)-2m\cdots \left(2\right)$

Solving equation $\left(1\right)$ and $\left(2\right)$, we get
$m(x+1)+\frac{1}{m}=-\frac{1}{m}(x+2)-2m\Rightarrow \frac{3}{m}+x(m+\frac{1}{m})+3m=0\Rightarrow (m+\frac{1}{m})(x+3)=0$
Since $(m+\frac{1}{m})\ne 0$, we get
$x=-3$