MCQ
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 $4 \sqrt{2}(\mathrm{~m}+\mathrm{c})$ is equal to $.....$
- ✓$34$
- B$64$
- C$62$
- D$32$
focus: $(-16,0)$
$\mathrm{y}=\mathrm{mx}+\mathrm{c}$ is focal chord
$\Rightarrow \mathrm{c}=16 \mathrm{~m} \ldots \ldots$. $(1)$
$y=m x+c$ is tangent to $(x+10)^{2}+y^{2}=4$
$\Rightarrow y-m(x+10) \pm 2 \sqrt{1+m^{2}}$
$\Rightarrow \mathrm{c}=10 \mathrm{~m} \pm 2 \sqrt{1+\mathrm{m}^{2}}$
$\Rightarrow 16 \mathrm{~m}=10 \pm 2 \sqrt{1+m^{2}}$
$\Rightarrow 6 m=2 \sqrt{1+m^{2}}(m>0)$
$\Rightarrow 9 \mathrm{~m}^{2}=1+\mathrm{m}^{2}$
$\Rightarrow m=\frac{1}{2 \sqrt{2}} \& c=\frac{8}{\sqrt{2}}$
$4 \sqrt{2}(m+c)=4 \sqrt{2}\left(\frac{17}{2 \sqrt{2}}\right)=34$
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