Solution:
The given answer \( 61 \) can be realized only when the equation of directrix is taken to be \( x=\pm \frac{4}{\sqrt{13}} \)
Instead of \( x=\pm \frac{4}{\sqrt{3}} \)
\( \displaystyle 2\frac{b^{2}}{a}=9 \) and \( \displaystyle \frac{a}{e}=\frac{4}{\sqrt{13}} \)
\( \Rightarrow b^{2}=\frac{9}{2}a \) and \( e^{2}=\frac{13}{16}a^{2} \)
\( e^{2}=1+\frac{b^{2}}{a^{2}} \)
\( \displaystyle \frac{13}{16}a^{2}=1+\frac{9}{2}a\cdot \frac{1}{a^2}\)
\( \displaystyle \frac{13}{16}a^{2}=a+\frac{9}{2a} \)
\( \displaystyle {13}a^{2}=1+\frac{9}{2a} \)
\(\displaystyle 13a^{2}=16a+\frac{72}{2} \)
\( 13a^{2}-16a-72=0 \)
\( (a-2)(13a^{2}+26a+36)=0 \)
Notice \( 13a^{2}+26a+36=0 \)
has no real root
\( \Rightarrow a=2 \)
\( e=\frac{\sqrt{13}}{2} \Rightarrow b^{2}=a^{2}(e^{2}-1)=4\left(\frac{9}{4}\right)=9 \)
So, the hyperbola is
\( \displaystyle \frac{x^{2}}{4}-\frac{y^{2}}{9}=1 \)
\( \sqrt{3}x-y=\sqrt{3} \)
The tangent is
\( y-\sqrt{3}x+\sqrt{3}=0 \)
Whose slope is \( \sqrt{3} \)
Say it touches the hyperbola at \( (x_{0},y_{0}) \)
\( \displaystyle \frac{x\,x_{0}}{4}-\frac{y.\,y_{0}}{9}=1 \) and the tangent is \( \displaystyle x-\frac{y}{\sqrt{3}}=1 \Rightarrow \frac{x_{0}}{4}=1;\; \frac{y_{0}}{9}=\frac{1}{\sqrt{3}} \)
\( x_{0}=4;\; y_{0}=3\sqrt{3} \)
Focus \( \left(\pm \sqrt{13},0\right) \)
\( \displaystyle m=\sqrt{\left(4+\sqrt{13}\right)^{2}+27}\;\sqrt{\left(4-\sqrt{13}\right)^{2}+27} \) \(=\sqrt{(3)^2 + (27)^2 + 27 \left( \left( 4 + \sqrt{13} \right)^2 + \left( 4 - \sqrt{13} \right)^2 \right)} \)
\( \displaystyle =\sqrt{9+(27)^2+ 27\cdot2\cdot (29)}= 3\sqrt{1 + (81) + 3 \cdot 2 \cdot 29}=3\sqrt{256}=48 \)
So, \( 4e^{2}+m=13+48=61 \)

JEE MainsJEE AdvanceBITSATJEE Mains | PYQ

JEE Mains JEE Advance BITSAT
JEE Mains | PYQ