JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
Let \(e_1\) and \(e_2\) be the eccentricities of the ellipse \(\frac{\mathrm{x}^2}{\mathrm{~b}^2}+\frac{\mathrm{y}^2}{25}=1 \quad\) and the hyperbola \(\quad \frac{\mathrm{x}^2}{16}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\), respectively. If \(\mathrm{b} \lt 5\) and \(\mathrm{e}_1 \mathrm{e}_2=1\), then the eccentricity of the ellipse having its axes along the coordinate axes and passing through all four foci (two of the ellipse and two of the hyperbola) is :
- A \(\frac{4}{5}\)
- B \(\frac{3}{5}\)
- C \(\frac{\sqrt{7}}{4}\)
- D \(\frac{\sqrt{3}}{2}\)
Answer & Solution
Correct Answer
(B) \(\frac{3}{5}\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \mathrm{e}_1^2=1-\frac{\mathrm{b}^2}{25} \quad \mathrm{e}_2^2=1-\frac{\mathrm{b}^2}{16} \\ & \therefore \mathrm{e}_1^2 \mathrm{e}_2^2=1 \\ & \left(1-\frac{\mathrm{b}^2}{25}\right)\left(1+\frac{\mathrm{b}^2}{16}\right)=1 \\ & \Rightarrow…
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