Let y=mx+c, m0 be the focal chord of y^2=-64x, which is tangent to (x+10)^2+y^2=4.

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 $4\sqrt2(m+c)$ is equal to \_\_\_\_\_\_\_\_.

Answer: 34

Accepted Answer

$y^2=4ax$ with $a=-16$; focus $(-16,0)$: $0=-16m+c\Rightarrow c=16m$.

Tangency to circle centre $(-10,0)$, $r=2$: $\dfrac{|-10m+c|}{\sqrt{m^2+1}}=2\Rightarrow6|m|=2\sqrt{m^2+1}\Rightarrow9m^2=m^2+1\Rightarrow m^2=\dfrac18$.

$$m=\frac1{2\sqrt2},\qquad c=16m=4\sqrt2.$$

$$m+c=\frac1{2\sqrt2}+4\sqrt2=\frac{\sqrt2}4+\frac{16\sqrt2}4=\frac{17\sqrt2}4.$$

$$4\sqrt2(m+c)=4\sqrt2\cdot\frac{17\sqrt2}4=34.$$

Solutions · 1