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JEE Main 202523 JanuaryMorning Shift
Exam
JEE Main
Year
2025
Date
23 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Hyperbola

JEE Main 2025 · 23 January · Morning Shift · Mathematics · Hyperbola

Let the circle C touch the line x-y+1=0 , have the centre on the...

Let the circle CC touch the line x−y+1=0x-y+1=0, have the centre on the positive x -axis, and cut off a chord of length 413\frac{4}{\sqrt{13}} along the line −3x+2y=1-3 x+2 y=1. Let H be the hyperbola x2α2−y2β2=1\frac{x^2}{\alpha^2}-\frac{y^2}{\beta^2}=1, whose one of the foci is the centre of CC and the length of the transverse axis is the diameter of CC. Then 2α2+3β22 \alpha^2+3 \beta^2 is equal to ______

Correct answer

0

Step-by-step solution

r=∣a+12∣⇒(a+1)2=2r2r=\left|\frac{a+1}{\sqrt{2}}\right| \Rightarrow(a+1)^2=2 r^2 Also (3a−113)2+(213)2=r2⇒(3a−113)2+413=(a+1)225a2−14a−3=0∴a=−15,3\left(\frac{3 a-1}{\sqrt{13}}\right)^2+\left(\frac{2}{\sqrt{13}}\right)^2=r^2 \begin{gathered} \Rightarrow\left(\frac{3 a-1}{\sqrt{13}}\right)^2+\frac{4}{13}=\frac{(a+1)^2}{2} \\ 5 a^2-14 a-3=0 \end{gathered} \therefore \quad a=-\frac{1}{5}, 3∵a≠−15⇒⇒r=22\begin{aligned} & \because \quad a \neq-\frac{1}{5} \Rightarrow \\ & \Rightarrow \quad r=2 \sqrt{2} \end{aligned}∵\because One focus of x2α2−y2β2=1\frac{x^2}{\alpha^2}-\frac{y^2}{\beta^2}=1 is (3,0)(3,0)⇒αe=3 and  2α=42⇒α=22⇒α2=8α2[1+β2α2]=9α2+β2=9⇒β2=1∴2α2+3β2=19\begin{aligned} & \Rightarrow \alpha e=3 \text{\text{ and } } 2 \alpha=4 \sqrt{2} \\ & \Rightarrow \alpha=2 \sqrt{2} \Rightarrow \alpha^2=8 \\ & \alpha^2\left[1+\frac{\beta^2}{\alpha^2}\right]=9 \\ & \alpha^2+\beta^2=9 \\ & \Rightarrow \beta^2=1 \\ & \therefore 2 \alpha^2+3 \beta^2=19\end{aligned}$$
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