If the focal chord of the hyperbola subtends | vedclass

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(C) Let the hyperbola be $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. The focus is at $(ae, 0)$.Since the focal chord is perpendicular to the transverse a

Vedclass · Accepted Answer

$e=\frac{\sqrt{5}+1}{2}$

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(C) Let the hyperbola be $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. The focus is at $(ae, 0)$.Since the focal chord is perpendicular to the transverse axis,its equation is $x = ae$.Substituting $x = ae$ in the hyperbola equation: $\frac{a^2e^2}{a^2} - \frac{y^2}{b^2} = 1$ $\Rightarrow e^2 - 1 = \frac{y^2}{b^2}$ $\Rightarrow y^2 = b^2(e^2 - 1)$.Since $e^2 = 1 + \frac{b^2}{a^2}$,we have $e^2 - 1 = \frac{b^2}{a^2}$,so $y^2 = b^2(\frac{b^2}{a^2}) = \frac{b^4}{a^2} \Rightarrow y = \pm \frac{b^2}{a}$.The endpoints of the focal chord are $A(ae, \frac{b^2}{a})$ and $B(ae, -\frac{b^2}{a})$.The center is $O(0, 0)$. The chord subtends a right angle at the center,so the slope of $OA$ multiplied by the slope of $OB$ is $-1$.Slope of $OA = \frac{b^2/a}{ae} = \frac{b^2}{a^2e}$. Slope of $OB = \frac{-b^2/a}{ae} = -\frac{b^2}{a^2e}$.$(\frac{b^2}{a^2e}) 	imes (-\frac{b^2}{a^2e}) = -1$ $\Rightarrow \frac{b^4}{a^4e^2} = 1$ $\Rightarrow b^4 = a^4e^2$.Since $b^2 = a^2(e^2 - 1)$,we have $(a^2(e^2 - 1))^2 = a^4e^2 \Rightarrow (e^2 - 1)^2 = e^2$.$e^4 - 2e^2 + 1 = e^2 \Rightarrow e^4 - 3e^2 + 1 = 0$.Let $u = e^2$. Then $u^2 - 3u + 1 = 0$. Using the quadratic formula: $u = \frac{3 \pm \sqrt{9 - 4}}{2} = \frac{3 \pm \sqrt{5}}{2}$.Since $e > 1$,$e^2 > 1$. $\frac{3 - \sqrt{5}}{2} \approx 0.38 $e^2 = \frac{6 + 2\sqrt{5}}{4} = \frac{(\sqrt{5} + 1)^2}{4} \Rightarrow e = \frac{\sqrt{5} + 1}{2}$.

If the focal chord of the hyperbola subtends a right angle at the center,then its eccentricity is

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