Question

Let $y=mx+c,m>0$ be the focal chord of $y^2=-64x,$ which is tangent to $(x+10{)}^2+y^2=4.$ Then the value of $4\sqrt{2}(m+c)$ is equal to

Answer 1

For $y^2=-64x$ focus is $(-16,0)$
$\because y=mx+c$ passes through $(–16,0)$
then $c=16m\dots \left(1\right)$
Also $y=mx+c$ touches the given circle
So, $\left|\frac{-10m+c}{\sqrt{1+m^2}}\right|=2$
From equation $\left(1\right)$
$\Rightarrow |3m|=\sqrt{1+m^2}$
$\Rightarrow m=\frac{1}{2\sqrt{2}}$ and $c=4\sqrt{2}$
Now, $4\sqrt{2}(m+c)=2\cdot 17=34$