JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
Let the ellipse \(E: \frac{x^{2}}{144}+\frac{y^{2}}{169}=1\) and the hyperbola \(H:\frac{x^{2}}{16}-\frac{y^{2}}{\lambda^{2}}=-1\) have the same foci. If e and L respectively denote the eccentricity and the length of the latus rectum of H, then the value of \(24(e+L)\) is:
- A 296
- B 126
- C 148
- D 67
Answer & Solution
Correct Answer
(A) 296
Step-by-step Solution
Detailed explanation
Equation of hyperbola : \(\frac{y^{2}}{\lambda^{2}}-\frac{x^{2}}{16}=1\) Equation of ellipse: \(\frac{x^{2}}{144}+\frac{y^{2}}{169}=1\) \(e^{\prime}=\sqrt{1-\frac{144}{169}}=\frac{5}{13}\) focus \(\Rightarrow(0,5)\) \(\Rightarrow \lambda \sqrt{1+\frac{16}{\lambda^2}}=5\)…
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