Question

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

Answer 1

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