JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
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 \(.....\)
- A \(34\)
- B \(64\)
- C \(62\)
- D \(32\)
Answer & Solution
Correct Answer
(A) \(34\)
Step-by-step Solution
Detailed explanation
\(y^2=-64\) 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 \((x+10)^{2}+y^{2}=4\) \(\Rightarrow y-m(x+10) \pm 2 \sqrt{1+m^{2}}\)…
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