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

Answer: 34

$y^2=-64x$
focus : $(-16,0)$
$y=mx+c$ is focal chord
$\Rightarrow c=16m\dots \dots ..(1)$
$y=mx+c$ is tangent to $(x+10{)}^2+y^2=4$
$\Rightarrow y=m(x+10)\pm 2\sqrt{1+m^2}$
$\Rightarrow c=10m\pm 2\sqrt{1+m^2}$
$\Rightarrow 16m=10m\pm 2\sqrt{1+m^2}$
$\Rightarrow 6m=2\sqrt{1+m^2}(m>0)$
$\Rightarrow 9m^2=1+m^2$
$\Rightarrow m=\frac{1}{2\sqrt{2}}\&c=\frac{8}{\sqrt{2}}$
$4\sqrt{2}(m+c)=4\sqrt{2}\left(\frac{17}{2\sqrt{2}}\right)=34$