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JEE Main 20244 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
4 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Hyperbola

JEE Main 2024 · 4 April · Evening Shift · Mathematics · Hyperbola

Consider a hyperbola H having centre at the origin and foci on the...

Consider a hyperbola H\mathrm{H} having centre at the origin and foci on the x\mathrm{x}-axis. Let C1\mathrm{C}_1 be the circle touching the hyperbola H\mathrm{H} and having the centre at the origin. Let C2\mathrm{C}_2 be the circle touching the hyperbola H\mathrm{H} at its vertex and having the centre at one of its foci. If areas (in sq units)ofC1of C_1 and C2C_2 are 36π36 \pi and 4π4 \pi, respectively, then the length (in units) of latus rectum of H\mathrm{H} is

Options

  1. A143\frac{14}{3}
  2. B283\frac{28}{3}
  3. C113\frac{11}{3}
  4. D103\frac{10}{3}

Correct answer

B. 283\frac{28}{3}

Step-by-step solution

Let H:x2a2−y2 b2=1( b2=a2(e2−1))\mathrm{H}: \frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1 \quad\left(\mathrm{~b}^2=\mathrm{a}^2\left(\mathrm{e}^2-1\right)\right)∴eqn of  C1 Ar. =36ππa2=36πa=6\begin{aligned} & \therefore \mathrm{eq}^{\mathrm{n}} \text{\text{ of } } \mathrm{C}_1 \\ & \text{ Ar. }=36 \pi \\ & \pi \mathrm{a}^2=36 \pi \\ & \mathrm{a}=6 \end{aligned}∴eqn of C1=x2+y2=a2\therefore \mathrm{eq}^{\mathrm{n}} \text{ of } \mathrm{C}_1=\mathrm{x}^2+\mathrm{y}^2=\mathrm{a}^2
Now radius of C2\mathrm{C}_2 can be a(e−1)\mathrm{a}(\mathrm{e}-1) or a(e+1)\mathrm{a}(\mathrm{e}+1) for r=a(e−1)\mathrm{r}=\mathrm{a}(\mathrm{e}-1) for r=a(e+1)\mathrm{r}=\mathrm{a}(\mathrm{e}+1)
Ar. =4ππr2=4ππa2(e−1)2=4πa2(e+1)2=436π(e−1)2=4π36(e+1)2=4e−1=13e+1=13e=43−23=4 \pi \pi r^2=4 \pi \pi \mathrm{a}^2(\mathrm{e}-1)^2=4 \pi \mathrm{a}^2(\mathrm{e}+1)^2=4 36 \pi(e-1)^2=4 \pi 36(e+1)^2=4 \mathrm{e}-1=\frac{1}{3} \mathrm{e}+1=\frac{1}{3} \mathrm{e}=\frac{4}{3} -\frac{2}{3} Not possible ∴b2=36(169−1)=28∴LR=2 b2a=2×286=283\begin{aligned} & \therefore \mathrm{b}^2=36\left(\frac{16}{9}-1\right)=28 \\ & \therefore L R=\frac{2 \mathrm{~b}^2}{\mathrm{a}}=\frac{2 \times 28}{6}=\frac{28}{3} \end{aligned}
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