The vertices of a hyperbola H are (pm 6, 0) a | vedclass

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Question

(B) Given the vertices of the hyperbola are $(\pm 6, 0)$,we have $a = 6$. Given eccentricity $e = \frac{\sqrt{5}}{2}$. We know $b^2 = a^2(e^2 - 1) = 3

Vedclass · Accepted Answer

$216$

Vedclass · Answer

(B) Given the vertices of the hyperbola are $(\pm 6, 0)$,we have $a = 6$. Given eccentricity $e = \frac{\sqrt{5}}{2}$. We know $b^2 = a^2(e^2 - 1) = 36 \left( \frac{5}{4} - 1 ight) = 36 \left( \frac{1}{4} ight) = 9$. So,the equation of the hyperbola $H$ is $\frac{x^2}{36} - \frac{y^2}{9} = 1$. The equation of the normal to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ at point $(a \sec 	heta, b 	an 	heta)$ is $\frac{ax}{\sec 	heta} + \frac{by}{	an 	heta} = a^2 + b^2$. Substituting $a=6, b=3$,the equation is $\frac{6x}{\sec 	heta} + \frac{3y}{	an 	heta} = 36 + 9 = 45$,which simplifies to $6x \cos 	heta + 3y \cot 	heta = 45$. The slope of this normal is $-\frac{6 \cos 	heta}{3 \cot 	heta} = -2 \sin 	heta$. The normal is parallel to $\sqrt{2}x + y = 2\sqrt{2}$,which has a slope of $-\sqrt{2}$. So,$-2 \sin 	heta = -\sqrt{2} \Rightarrow \sin 	heta = \frac{1}{\sqrt{2}} \Rightarrow 	heta = \frac{\pi}{4}$. Substituting $	heta = \frac{\pi}{4}$ into the normal equation: $6x \cos(\frac{\pi}{4}) + 3y \cot(\frac{\pi}{4}) = 45 \Rightarrow 6x(\frac{1}{\sqrt{2}}) + 3y(1) = 45 \Rightarrow 3\sqrt{2}x + 3y = 45 \Rightarrow \sqrt{2}x + y = 15$. The point $P$ on the hyperbola is $(6 \sec(\frac{\pi}{4}), 3 	an(\frac{\pi}{4})) = (6\sqrt{2}, 3)$. The intersection of the normal with the $y$-axis $(x=0)$ is $K(0, 15)$. The length $d$ is the distance between $P(6\sqrt{2}, 3)$ and $K(0, 15)$. $d^2 = (6\sqrt{2} - 0)^2 + (3 - 15)^2 = (6\sqrt{2})^2 + (-12)^2 = 72 + 144 = 216$.

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