JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
Let \(e_{1}\) and \(e_{2}\) be the eccentricities of the ellipse, \(\frac{x^{2}}{25}+\frac{y^{2}}{b^{2}}=1(b<5)\) and the hyperbola \(\frac{ x ^{2}}{16}-\frac{ y ^{2}}{ b ^{2}}=1\) respectively satisfying \(e _{1} e _{2}=1 .\) If \(\alpha\) and \(\beta\) are the distances between the foci of the ellipse and the foci of the hyperbola respectively, then the ordered pair \((\alpha, \beta)\) is equal to
- A \((8,10)\)
- B \((8,12)\)
- C \(\left(\frac{20}{3}, 12\right)\)
- D \(\left(\frac{24}{5}, 10\right)\)
Answer & Solution
Correct Answer
(A) \((8,10)\)
Step-by-step Solution
Detailed explanation
For ellipse \(\frac{x^{2}}{25}+\frac{y^{2}}{b^{2}}=1 \quad(b<5)\) Let e \(_{1}\) is eccentricity of ellipse \(\therefore \quad b^{2}=25\left(1-e_{1}^{2}\right) \ldots \ldots .(1)\) Again for hyperbola \(\frac{x^{2}}{16}-\frac{y^{2}}{b^{2}}=1\) Let \(e _{2}\) is eccentricity of…
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