Thank you for reporting, we will resolve it shortly
Q.
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
Conic Sections
Solution:
Any tangent to $y^{2}= 4\left(x+1\right)$ is
$y=m\left(x+1\right)+\frac{1}{m}\quad...\left(1\right)$
Any tangent to $y^{2} = 8\left(x+2\right)$is
$ y=m'\left(x+2\right)+\frac{2}{m}\quad...\left(2\right) $
Since $\left(1\right)$ and $\left(2\right)$ are $\bot$
$ \therefore mm' = -1 $
$ \therefore m=-\frac{1}{m} $
Putting in $\left(2\right)$, we get
$ y = -\frac{1}{m}\left(x+2\right)-2m\quad...\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 \quad\left(\because m+\frac{1}{m}\ne0\right)$