JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
Let \(e_1\) be the eccentricity of the hyperbola \(\frac{x^2}{16}-\frac{y^2}{9}=1\) and \(e_2\) be the eccentricity of the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a>b\), which passes through the foci of the hyperbola. If \(e_1 e_2=1\), then the length of the chord of the ellipse parallel to the \(\mathrm{x}\)-axis and passing through \((0,2)\) is :
- A \(4 \sqrt{5}\)
- B \(\frac{8 \sqrt{5}}{3}\)
- C \(\frac{10 \sqrt{5}}{3}\)
- D \(3 \sqrt{5}\)
Answer & Solution
Correct Answer
(C) \(\frac{10 \sqrt{5}}{3}\)
Step-by-step Solution
Detailed explanation
\( H: \frac{x^2}{16}-\frac{y^2}{9}=1 \quad e_1=\frac{5}{4} \) \(\therefore e_1 e_2=1 \Rightarrow e_2=\frac{4}{5}\) Also, ellipse is passing through \(( \pm 5,0)\) \( \therefore a=5 \text { and } b=3 \) \( E: \frac{x^2}{25}+\frac{y^2}{9}=1\) End point of chord are…
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