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JEE Main 20257 AprilMorning Shift
Exam
JEE Main
Year
2025
Date
7 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Hyperbola

JEE Main 2025 · 7 April · Morning Shift · Mathematics · Hyperbola

Consider the hyperbola (x^2)/(a^2)-(y^2)/(b^2)=1 having one of its...

Consider the hyperbola x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 having one of its focus at P(−3,0)\mathrm{P}(-3,0). If the latus ractum through its other focus subtends a right angle at P and a2b2=α2−β,α,β∈Na^2 b^2=\alpha \sqrt{2}-\beta, \alpha, \beta \in \mathbb{N}.

Correct answer

1944

Step-by-step solution

f1≡(−ae,0)≡P(−3,0)⇒ae=3\begin{aligned} & \mathrm{f}_1 \equiv(-\mathrm{ae}, 0) \equiv \mathrm{P}(-3,0) \\ & \Rightarrow \mathrm{ae}=3\end{aligned}tan⁡45∘=b2/a2ae2ae=b2ab2=6a Also a2e2=a2+b29=a2+6aa2+6a−9=0a=−3±32=−3(1±2)∴a2b2=a2⋅6a=6a3=6(1352−189)α=810 and β=1134∴α+β=1944\begin{aligned} & \tan 45^{\circ}=\frac{b^2 / a}{2 a e} \\ & 2 a e=\frac{b^2}{a} \\ & b^2=6 a \\ & \text{ Also } a^2 e^2=a^2+b^2 \\ & 9=a^2+6 a \\ & a^2+6 a-9=0 \\ & a=-3 \pm 3 \sqrt{2}=-3(1 \pm \sqrt{2}) \\ & \therefore a^2 b^2=a^2 \cdot 6 a=6 a^3 \\ & =6(135 \sqrt{2}-189) \\ & \alpha=810 \text{ and } \beta=1134 \\ & \therefore \alpha+\beta=1944\end{aligned}$
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