JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
The area (in sq, units) of the quadrilateral formed by the tangents at the end points of the latera recta to the ellipse \(\frac{{{x^2}}}{9} + \frac{{{y^2}}}{5} = 1\) is :
- A \(27\)
- B \(\frac{{27}}{4}\)
- C \(18\)
- D \(\frac{{27}}{2}\)
Answer & Solution
Correct Answer
(A) \(27\)
Step-by-step Solution
Detailed explanation
\(a=3, b=\sqrt{5}\) \(e=\sqrt{1-\frac{5}{9}}=\frac{2}{3}\) foci \(=(\pm 2,0)\) tangent at \(P \Rightarrow \frac{2 x}{9}+\frac{5 y}{3.5}=1\) \(\frac{2 x}{9}+\frac{y}{3}=1\) \(2 x+3 y=9\) Area of quadrilateral \(=4 \times \text { (area of triangle } \mathrm{QCR})\)…
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