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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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 $K(0, 15)$. $d^2 = (6\sqrt{2} - 0)^2 + (3 - 15)^2 = (6\sqrt{2})^2 + (-12)^2 = 72 + 144 = 216$.

Similar Questions

If the transverse and conjugate axes of a hyperbola are equal,then its eccentricity is

The equation of the chord of the hyperbola $25x^{2} - 16y^{2} = 400$ whose midpoint is $(5, 3)$ is:

If the vertices of a hyperbola are at $(-2, 0)$ and $(2, 0)$ and one of its foci is at $(-3, 0)$,then which one of the following points does not lie on this hyperbola?

Let the foci of a hyperbola be $(1, 14)$ and $(1, -12)$. If it passes through the point $(1, 6)$,then the length of its latus-rectum is:

Let $e$ be the eccentricity of the ellipse $\frac{x^2}{4}+\frac{y^2}{9}=1$. If $\frac{1}{e}$ is the eccentricity of a hyperbola,then the eccentricity of its conjugate hyperbola is

Vedclass Test Series

Exam Paper Generator

Online Exam Module