JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
An ellipse \(E: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) passes through the vertices of the hyperbola \(H: \frac{x^{2}}{49}-\frac{y^{2}}{64}=-1\). Let the major and minor axes of the ellipse \(E\) coincide with the transverse and conjugate axes of the hyperbola \(H\). Let the product of the eccentricities of \(E\) and \(H\) be \(\frac{1}{2}\). If \(l\) is the length of the latus rectum of the ellipse \(E\), then the value of \(113\,l\) is equal to \(....\)
- A \(1500\)
- B \(1552\)
- C \(1000\)
- D \(1553\)
Answer & Solution
Correct Answer
(B) \(1552\)
Step-by-step Solution
Detailed explanation
Hyp : \(\frac{y^{2}}{64}-\frac{x^{2}}{49}=1\) An ellipse \(E: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) passes through the vertices of the hyperbola \(H: \frac{x^{2}}{49}-\frac{y^{2}}{64}=-1\). \(S_{0} b^{2}=64\) \(e_{H}=\sqrt{1+\frac{a^{2}}{b^{2}}}=\sqrt{1+\frac{49}{64}}\)…
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