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QJEEM24Ma2223: JEE Main 2024 Mathematics - Class 11 JEEM24Ma2223 Solution: The length of the latus rectum and directrices of hyperbola…

The length of the latus rectum and directrices of hyperbola with eccentricity e are 9 and x=±43x= \pm \frac{4}{\sqrt{3}} , respectively. Let the line y−3x+3=0y-\sqrt{3} x+\sqrt{3}=0 touch this hyperbola at (x0,y0)\left(x_0, y_0\right) . If m\mathrm{m} is the product of the focal distances of the point (x0,y0)\left(x_0, y_0\right) , then 4e2+m4 \mathrm{e}^2+\mathrm{m} is equal to _________.

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JEE Main 2024 (Online) 6th April Evening Shift Mathematics - Hyperbola Question 3 English Explanation

y=mx±a2m2−b2y=m x \pm \sqrt{a^2 m^2-b^2} is the tangent

amp;⇒m=3⇒a2m2−b2=3amp;⇒3a2−b2=3\begin{aligned}& \Rightarrow m=\sqrt{3} \Rightarrow a^2 m^2-b^2=3 \\& \Rightarrow 3 a^2-b^2=3\end{aligned}

amp;2b2a=9⇒b2=9a2⇒3a2−9a2=3amp;⇒a2−32a−1=0amp;⇒a=2 or −0.5 (ignore) amp;⇒b=3amp;⇒x24−y29=1amp;⇒ Solving hyperbola and tangent y=3x−3amp;x0=4,y0=33,e=1+94=132amp;PF1⋅PF2=amp;(4−13)2+(33)2(4+13)2+(33)2amp;=(56−813)(56+813)amp;=2304=48=mamp;⇒4e2+m=13+48=61\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}

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