JEE Main 2024 — Mathematics Coordinate Geometry

$\begin{aligned} & \mathrm{H}: \frac{\mathrm{y}^2}{\mathrm{~b}^2}-\frac{\mathrm{x}^2}{\mathrm{a}^2}=1, \mathrm{e}=\sqrt{3} \\ & \mathrm{e}=\sqrt{1+\frac{\mathrm{a}^2}{\mathrm{~b}^2}}=\sqrt{3} \Rightarrow \frac{\mathrm{a}^2}{\mathrm{~b}^2}=2 \\ & \qquad \mathrm{a}^2=2 \mathrm{~b}^2 \\ & \text { length of L.R. }=\frac{2 \mathrm{a}^2}{\mathrm{~b}}=4 \sqrt{3} \\ & \qquad a=\sqrt{6}\end{aligned}$​H: b2y2​−a2x2​=1,e=3​e=1+ b2a2​​=3​⇒ b2a2​=2a2=2 b2 length of L.R. = b2a2​=43​a=6​​ $P(\alpha, 6)$P(α,6) lie on $\frac{y^2}{3}-\frac{x^2}{6}=1$3y2​−6x2​=1 $\begin{aligned}
& 12-\frac{\alpha^2}{6}=1 \Rightarrow \alpha^2=66 \
& \text { Foci }=(0, \pm \text { be })=(0,3) &(0,-3)
\end{aligned}$Let$Letd_1 & d_2$be focal distances of$befocaldistancesof\mathrm{P}(\alpha, 6)$$\begin{aligned} & \mathrm{d}_1=\sqrt{\alpha^2+(6+b e)^2}, \mathrm{d}_2=\sqrt{\alpha^2+(6-b e)^2} \ & \mathrm{d}_1=\sqrt{66+81}, \mathrm{d}_2=\sqrt{66+9} \ & \beta=\mathrm{d}_1 \mathrm{d}_2=\sqrt{147 \times 75}=105 \ & \alpha^2+\beta=66+105=171\end{aligned}$

SECTION - B