TS EAMCET · Maths · Ellipse
Tangents are drawn to the ellipse \(\frac{x^2}{9}+\frac{y^2}{5}=1\) at all the ends of its latus rectum. The area of the quadrilateral so formed (in sq. units) is
- A \(27\)
- B \(36\)
- C \(42\)
- D \(45\)
Answer & Solution
Correct Answer
(A) \(27\)
Step-by-step Solution
Detailed explanation
Ends of latus rectum of \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) \(\left( \pm a e, \pm b \sqrt{1-e^2}\right)\) \[ \begin{aligned} & \frac{x^2}{9}+\frac{y^2}{5}=1 \\ & a=3, b=\sqrt{5}, e^2=1-\frac{5}{9}=\frac{4}{9} \end{aligned} \] \(\therefore\) Ends of latus rectum :…
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