TS EAMCET · Maths · Ellipse
The area (in sq. units) of the quadrilateral formed by the tangents drawn at the end points of the latus rectum to the ellipse \(S \equiv \frac{x^2}{16}+\frac{y^2}{12}=1\) is
- A 96
- B 16
- C 128
- D 64
Answer & Solution
Correct Answer
(D) 64
Step-by-step Solution
Detailed explanation
Given ellipse, \(\frac{x^2}{16}+\frac{y^2}{12}=1\) End point of latus rectum \(=\left( \pm a e, \frac{2 b^2}{a}\right)=(2,3)\) Equation of tangent at \((2,3)\) of ellipse \(\frac{x^2}{16}+\frac{y^2}{12}=1\) is,…
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