JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
Let for two distinct values of \(p\) the lines \(y=x+p\) touch the ellipse \(\mathrm{E}: \frac{\mathrm{x}^2}{4^2}+\frac{\mathrm{y}^2}{3^2}=1\) at the points A and B . Let the line \(\mathrm{y}=\mathrm{x}\) intersect E at the points C and \(D\). Then the area of the quadrilateral \(A B C D\) is equal to
- A \(36\)
- B \(24\)
- C \(48\)
- D \(20\)
Answer & Solution
Correct Answer
(B) \(24\)
Step-by-step Solution
Detailed explanation
Point of contact are \(\left(\frac{\mp \mathrm{a}^2 \mathrm{~m}}{\sqrt{\mathrm{a}^2 \mathrm{~m}^2+\mathrm{b}^2}}, \frac{ \pm \mathrm{b}^2}{\sqrt{\mathrm{a}^2 \mathrm{~m}^2+\mathrm{b}^2}}\right)\)…
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