mages/batch-001/3be115a711ac.webp" data-orsrc="https://uploads-pracbee-ramanuj.s3.ap-south-1.amazonaws.com/pyq-question-images/batch-001/3be115a711ac.webp" loading="lazy" style="max-width:100%;height:auto;display:block;margin:0 auto;max-height:40vh;vertical-align:baseline" alt="JEE Main 2024 (Online) 6th April Evening Shift Mathematics - Hyperbola Question 23 English Explanation">

$$y=m x \pm \sqrt{a^2 m^2-b^2}$$ is the tangent

$$\begin{aligned} & \Rightarrow m=\sqrt{3} \Rightarrow a^2 m^2-b^2=3 \\ & \Rightarrow 3 a^2-b^2=3 \end{aligned}$$

$$\begin{aligned} & \frac{2 b^2}{a}=9 \Rightarrow b^2=\frac{9 a}{2} \Rightarrow 3 a^2-\frac{9 a}{2}=3 \\ & \Rightarrow a^2-\frac{3}{2} a-1=0 \\ & \Rightarrow a=2 \text { or }-0.5 \text { (ignore) } \\ & \Rightarrow b=3 \\ & \Rightarrow \frac{x^2}{4}-\frac{y^2}{9}=1 \\ & \Rightarrow \text { Solving hyperbola and tangent } y=\sqrt{3 x}-\sqrt{3} \\ & x_0=4, y_0=3 \sqrt{3}, e=\sqrt{1+\frac{9}{4}}=\frac{\sqrt{13}}{2} \\ & P F_1 \cdot P F_2= \\ & \sqrt{(4-\sqrt{13})^2+(3 \sqrt{3})^2} \sqrt{(4+\sqrt{13})^2+(3 \sqrt{3})^2} \\ & =\sqrt{(56-8 \sqrt{13})(56+8 \sqrt{13})} \\ & =\sqrt{2304}=48=m \\ & \Rightarrow 4 e^2+m=13+48=61 \end{aligned}$$

19:T5b3,

JEE Main 2024 (Online) 6th April Evening Shift Mathematics - Hyperbola Question 23 Hindi Explanation

$$y=m x \pm \sqrt{a^2 m^2-b^2}$$ स्पर्श रेखा है

$$\begin{aligned} & \Rightarrow m=\sqrt{3} \Rightarrow a^2 m^2-b^2=3 \\ & \Rightarrow 3 a^2-b^2=3 \end{aligned}$$

$$\begin{aligned} & \frac{2 b^2}{a}=9 \Rightarrow b^2=\frac{9 a}{2} \Rightarrow 3 a^2-

Hyperbola

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