TS EAMCET · Maths · Ellipse
Let e be the eccentricity of the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\). If \(a=5, b=4\) and the equation of the normal drawn at one end of the latus rectum that lies in the first quadrant is \(l x+m y=27\), then \(l+m=\)
- A \(\frac{3}{e}\)
- B \(\frac{3}{2 e}\)
- C \(\frac{6}{e}\)
- D \(\frac{1}{e}\)
Answer & Solution
Correct Answer
(C) \(\frac{6}{e}\)
Step-by-step Solution
Detailed explanation
\(e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{4^2}{5^2}} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}\) End of latus rectum in 1st quadrant: \((ae, \frac{b^2}{a}) = (5 \cdot \frac{3}{5}, \frac{4^2}{5}) = (3, \frac{16}{5})\) Equation of normal:…
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