WBJEE · Maths · Ellipse
Tangents are drawn to the ellipse \(\frac{x^{2}}{9}+\frac{y^{2}}{5}=1\) at the ends of both latusrectum. The area of the quadrilateral, so formed is
- A 27 sq units
- B \(\frac{13}{2}\) sq units
- C \(\frac{15}{4} \mathrm{sq}\) units
- D 45 sq units
Answer & Solution
Correct Answer
(A) 27 sq units
Step-by-step Solution
Detailed explanation
We have, \(\frac{x^{2}}{9}+\frac{y^{2}}{5}=1\) \(\Rightarrow \quad a=3\) and \(b=\sqrt{5}\) \(\begin{aligned} \therefore \quad e &=\sqrt{1-\frac{b^{2}}{a^{2}}} \\ &=\sqrt{1-\frac{5}{9}}=\frac{2}{3} \end{aligned}\) \(\therefore\) Foci \(=(\pm a e, 0)\)…
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