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ધારોકે $H: \frac{-x^2}{a^2}+\frac{y^2}{b^2}=1$ અતિવલય છે, જેની ઉત્કેન્દ્રતા $\sqrt{3}$ અને નાભીલંબની લંબાઈ $4 \sqrt{3}$ છે. ધારોકે $(\alpha, 6), \alpha>0$ એ $H$ પર છે. જો બિંદુ ( $\alpha, 6)$ ના નાભ્યાંતરોનો ગુણાકાર $\beta$ હોય, તો $\alpha^2+\beta=$............
$170$
$171$
$169$
$172$
Solution
$ \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} \quad \Rightarrow \frac{\mathrm{a}^2}{\mathrm{~b}^2}=2 $
$ a^2=2 b^2 $
$ \text { length of L.R. }=\frac{2 a^2}{b}=4 \sqrt{3} $
$ \mathrm{a}=\sqrt{6} $
$ P(\alpha, 6) \text { lie on } \frac{y^2}{3}-\frac{x^2}{6}=1 $
$ 12-\frac{\alpha^2}{6}=1 \Rightarrow \alpha^2=66 $
$ \text { Foci }=(0, \pm \mathrm{be})=(0,3) \&(0,-3) $
Let $d_1 \& d_2$ be focal distances of $\mathrm{P}(\alpha, 6)$
$ d_1=\sqrt{\alpha^2+(6+b e)^2}, d_2=\sqrt{\alpha^2+(6-b e)^2} $
$ d_1=\sqrt{66+81}, d_2=\sqrt{66+9} $
$ \beta=d_1 d_2=\sqrt{147 \times 75}=105 $
$ \alpha^2+\beta=66+105=171$
