Consider the hyperbola frac{x^2}{a^2}-frac{y^ | vedclass

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(C) Let the foci be $F_1(-ae, 0)$ and $F_2(ae, 0)$. Given $F_1 = P(-3, 0)$,so $ae = 3$.The latus rectum passes through $F_2(ae, 0)$ and has endpoints

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$1944$

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(C) Let the foci be $F_1(-ae, 0)$ and $F_2(ae, 0)$. Given $F_1 = P(-3, 0)$,so $ae = 3$.The latus rectum passes through $F_2(ae, 0)$ and has endpoints $L_1(ae, b^2/a)$ and $L_2(ae, -b^2/a)$.The angle $\angle L_1 P L_2 = 90^\circ$. Since the triangle is symmetric about the x-axis,the angle $\angle L_1 P F_2 = 45^\circ$.In $	riangle L_1 P F_2$,$	an 45^\circ = \frac{L_1 F_2}{P F_2} = \frac{b^2/a}{2ae} = 1$.Thus,$b^2 = 2a(ae) = 2a(3) = 6a$.Using the hyperbola property $b^2 = a^2(e^2 - 1) = a^2e^2 - a^2$,we have $6a = 9 - a^2$,so $a^2 + 6a - 9 = 0$.Solving for $a$,$a = \frac{-6 \pm \sqrt{36 - 4(1)(-9)}}{2} = -3 \pm 3\sqrt{2}$. Since $a > 0$,$a = 3\sqrt{2} - 3$.Then $a^2 = (3\sqrt{2}-3)^2 = 18 + 9 - 18\sqrt{2} = 27 - 18\sqrt{2}$.$b^2 = 6a = 6(3\sqrt{2}-3) = 18\sqrt{2} - 18$.$a^2b^2 = (27 - 18\sqrt{2})(18\sqrt{2} - 18) = 486\sqrt{2} - 486 - 648 + 324\sqrt{2} = 810\sqrt{2} - 1134$.Comparing with $\alpha\sqrt{2} - \beta$,we get $\alpha = 810$ and $\beta = 1134$.Therefore,$\alpha + \beta = 810 + 1134 = 1944$.

Consider the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ having one of its foci at $P(-3,0)$. If the latus rectum through its other focus subtends a right angle at $P$ and $a^2b^2 = \alpha\sqrt{2} - \beta$,where $\alpha, \beta \in N$,then find the value of $\alpha + \beta$.

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