Let X-axis be the transverse axis and Y-axis | vedclass

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

Question

(C) The eccentricity $e'$ of the ellipse $\frac{x^2}{4} + \frac{y^2}{2} = 1$ is given by:$e' = \sqrt{1 - \frac{2}{4}} = \sqrt{1 - \frac{1}{2}} = \frac

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(C) The eccentricity $e'$ of the ellipse $\frac{x^2}{4} + \frac{y^2}{2} = 1$ is given by:$e' = \sqrt{1 - \frac{2}{4}} = \sqrt{1 - \frac{1}{2}} = \frac{1}{\sqrt{2}}$Given that the eccentricity $e$ of the hyperbola $H$ is the reciprocal of $e'$,we have:$e = \frac{1}{e'} = \sqrt{2}$For a hyperbola with transverse axis along the $X$-axis,$e = \sqrt{1 + \frac{b^2}{a^2}}$. Thus:$\sqrt{1 + \frac{b^2}{a^2}} = \sqrt{2} \implies 1 + \frac{b^2}{a^2} = 2 \implies \frac{b^2}{a^2} = 1 \implies b^2 = a^2$The equation of the hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{a^2} = 1$.Since the point $(5, 4)$ lies on $H$:$\frac{25}{a^2} - \frac{16}{a^2} = 1 \implies \frac{9}{a^2} = 1 \implies a^2 = 9 \implies a = 3$The length of the transverse axis is $2a = 2 	imes 3 = 6$.

Let $X$-axis be the transverse axis and $Y$-axis be the conjugate axis of a hyperbola $H$. Let the eccentricity of $H$ be the reciprocal of the eccentricity of the ellipse $\frac{x^2}{4} + \frac{y^2}{2} = 1$. If $(5, 4)$ is a point on $H$,then the length of the transverse axis of $H$ is

Similar Questions

The equation of the hyperbola with focus $(1, 2)$,eccentricity $e = \sqrt{3}$,and directrix $2x + y = 1$ is given by:

The length of the latus rectum of the curve $xy = 7x + 5y$ is

The equation of the tangent parallel to $y-x+5=0$ drawn to $\frac{x^{2}}{3}-\frac{y^{2}}{2}=1$ is

If $l_1$ and $l_2$ are the lengths of the perpendiculars drawn from a point on the hyperbola $5x^2 - 4y^2 - 20 = 0$ to its asymptotes,then $\frac{l_1^2 l_2^2}{100} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module