Let H: frac{-x^2}{a^2}+frac{y^2}{b^2}=1 be th | vedclass

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(B) Given the hyperbola $H: \frac{y^2}{b^2} - \frac{x^2}{a^2} = 1$ with eccentricity $e = \sqrt{3}$.For a vertical hyperbola,$e^2 = 1 + \frac{a^2}{b^2

Vedclass · Accepted Answer

$171$

Vedclass · Answer

(B) Given the hyperbola $H: \frac{y^2}{b^2} - \frac{x^2}{a^2} = 1$ with eccentricity $e = \sqrt{3}$.For a vertical hyperbola,$e^2 = 1 + \frac{a^2}{b^2}$,so $3 = 1 + \frac{a^2}{b^2}$ $\Rightarrow \frac{a^2}{b^2} = 2$ $\Rightarrow a^2 = 2b^2$.The length of the latus rectum is $\frac{2a^2}{b} = 4\sqrt{3}$.Substituting $a^2 = 2b^2$,we get $\frac{2(2b^2)}{b} = 4b = 4\sqrt{3}$,so $b = \sqrt{3}$ and $b^2 = 3$.Then $a^2 = 2(3) = 6$.The equation of the hyperbola is $\frac{y^2}{3} - \frac{x^2}{6} = 1$.Since $(\alpha, 6)$ lies on $H$,$\frac{6^2}{3} - \frac{\alpha^2}{6} = 1$ $\Rightarrow 12 - \frac{\alpha^2}{6} = 1$ $\Rightarrow \frac{\alpha^2}{6} = 11$ $\Rightarrow \alpha^2 = 66$.The foci are $(0, \pm be) = (0, \pm \sqrt{3} \cdot \sqrt{3}) = (0, \pm 3)$.The focal distances $d_1, d_2$ of a point $(x, y)$ on the hyperbola $\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1$ are $|ey \pm b|$.Here $e = \sqrt{3}, b = \sqrt{3}, y = 6$,so $d_1, d_2 = |\sqrt{3}(6) \pm \sqrt{3}| = |6\sqrt{3} \pm \sqrt{3}|$.$d_1 = 7\sqrt{3}$ and $d_2 = 5\sqrt{3}$.$\beta = d_1 d_2 = (7\sqrt{3})(5\sqrt{3}) = 35 \cdot 3 = 105$.Therefore,$\alpha^2 + \beta = 66 + 105 = 171$.

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