JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
Let the foci of a hyperbola \(\mathrm{H}\) coincide with the foci of the ellipse \(E: \frac{(x-1)^2}{100}+\frac{(y-1)^2}{75}=1\) and the eccentricity of the hyperbola \(\mathrm{H}\) be the reciprocal of the eccentricity of the ellipse \(E\). If the length of the transverse axis of \(\mathrm{H}\) is \(\alpha\) and the length of its conjugate axis is \(\beta\), then \(3 \alpha^2+2 \beta^2\) is equal to :
- A \(242\)
- B \(225\)
- C \(237\)
- D \(205\)
Answer & Solution
Correct Answer
(B) \(225\)
Step-by-step Solution
Detailed explanation
\( \mathrm{e}_1=\sqrt{1-\frac{75}{100}}=\frac{5}{10}=\frac{1}{2} \) \( \mathrm{e}_2=2 \) \( \mathrm{~F}_1(6,1), \mathrm{F}_2(-4,1) \) \( 2 \mathrm{ae}_2=10 \Rightarrow \mathrm{a}=\frac{5}{2} \Rightarrow 2 \mathrm{a}=5 \) \( \Rightarrow \alpha=5 \)…
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