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JEE Main 20249 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
9 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Hyperbola

JEE Main 2024 · 9 April · Evening Shift · Mathematics · Hyperbola

Let the foci of a hyperbola H coincide with the foci of the ellipse...

Let the foci of a hyperbola HH coincide with the foci of the ellipse E:(x−1)2100+(y−1)275=1E: \frac{(x-1)^2}{100}+\frac{(y-1)^2}{75}=1 and the eccentricity of the hyperbola HH be the reciprocal of the eccentricity of the ellipse EE. If the length of the transverse axis of HH is α\alpha and the length of its conjugate axis is β\beta, then 3α2+2β23 \alpha^2+2 \beta^2 is equal toLet the foci of a hyperbola H coincide with the foci of the ellipse E: ((x-1)^2)/(100)+((y-1)^2)/(75)=1 and the

Options

  1. A237
  2. B242
  3. C205
  4. D225

Correct answer

D. 225

Step-by-step solution

Let the foci of a hyperbola H coincide with the foci of the ellipse E: ((x-1)^2)/(100)+((y-1)^2)/(75)=1 and the
e1=1−75100=510=12e2=2 F1(6,1),F2(−4,1)2ae2=10⇒a=52⇒2a=5⇒α=54=1+b2a2⇒b2=3a2 b=3×52β=533α2+2β2=3×25+2×25×3=225\begin{aligned} & \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 \\ & 4=1+\frac{\mathrm{b}^2}{\mathrm{a}^2} \Rightarrow \mathrm{b}^2=3 \mathrm{a}^2 \\ & \mathrm{~b}=\sqrt{3} \times \frac{5}{2} \\ & \beta=5 \sqrt{3} \\ & 3 \alpha^2+2 \beta^2=3 \times 25+2 \times 25 \times 3 \\ & =225\end{aligned}
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