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QJEEM24Maa5a3: JEE Main 2024 Mathematics - Class 11 JEEM24Maa5a3 Solution: Consider a hyperbola having centre at the origin and foci on…

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) of C1C_1 and C2C_2 are 36π36 \pi and 4π4 \pi , respectively, then the length (in units) of latus rectum of H\mathrm{H} is

Multiple Choice Options

  • A 283\frac{28}{3}
  • B 113\frac{11}{3}
  • C 143\frac{14}{3}
  • D 103\frac{10}{3}

Step-by-Step Verified Solution

JEE Main 2024 (Online) 4th April Evening Shift Mathematics - Hyperbola Question 6 English Explanation

amp;C1:x2+y2=a2⇒ area =πa2=36π⇒a=6amp;C2:x2+(y−ae)2=(ae−a)2amp;∴π(ae−a)2=4πamp;⇒36(e−1)2=4amp;⇒e−1=13⇒e=43amp;⇒b2=28amp;∴LR=2b2a=2×286=283 units \begin{aligned}& C_1: x^2+y^2=a^2 \Rightarrow \text { area }=\pi a^2=36 \pi \Rightarrow a=6 \\& C_2: x^2+(y-a e)^2=(a e-a)^2 \\& \therefore \quad \pi(a e-a)^2=4 \pi \\& \Rightarrow 36(e-1)^2=4 \\& \Rightarrow e-1=\frac{1}{3} \Rightarrow e=\frac{4}{3} \\& \Rightarrow b^2=28 \\& \therefore \quad L R=\frac{2 b^2}{a}=\frac{2 \times 28}{6}=\frac{28}{3} \text { units }\end{aligned}

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