If the latus rectum of a hyperbola subtends a | vedclass

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(D) Let the equation of the hyperbola be $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.The coordinates of the focus are $(ae, 0)$ and the length of the semi

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$2$

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(D) Let the equation of the hyperbola be $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.The coordinates of the focus are $(ae, 0)$ and the length of the semi-latus rectum is $\frac{b^2}{a}$.The point at the end of the latus rectum is $(ae, \frac{b^2}{a})$.The slope of the line joining the center $(0, 0)$ and the point $(ae, \frac{b^2}{a})$ is $m_1 = \frac{b^2/a}{ae} = \frac{b^2}{a^2e}$.Since the latus rectum subtends a right angle at the center,the slopes of the lines to the two ends of the latus rectum must be $1$ and $-1$.Thus,$\frac{b^2}{a^2e} = 1$,which implies $b^2 = a^2e$.Using the relation $b^2 = a^2(e^2 - 1)$,we get $a^2(e^2 - 1) = a^2e$.$e^2 - 1 = e \implies e^2 - e - 1 = 0$.Solving for $e$ using the quadratic formula $e = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $e = \frac{1 \pm \sqrt{1 - 4(1)(-1)}}{2} = \frac{1 \pm \sqrt{5}}{2}$.Since $e > 1$,we take $e = \frac{1 + \sqrt{5}}{2}$,which is the golden ratio $\phi$.

If the latus rectum of a hyperbola subtends a right angle at the center,then its eccentricity is:

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