A hyperbola passes through a focus of the ell | vedclass

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

Question

(C) Let the equation of the hyperbola be $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. The given equation of the ellipse is $\frac{x^2}{13^2} + \frac{y^2}{

Vedclass · Accepted Answer

$\frac{x^2}{144}-\frac{y^2}{25}=1$

Vedclass · Answer

(C) Let the equation of the hyperbola be $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. The given equation of the ellipse is $\frac{x^2}{13^2} + \frac{y^2}{5^2} = 1$. For the ellipse,$a_e = 13$ and $b_e = 5$. The eccentricity of the ellipse $e$ is given by $e = \sqrt{1 - \frac{b_e^2}{a_e^2}} = \sqrt{1 - \frac{25}{169}} = \sqrt{\frac{144}{169}} = \frac{12}{13}$. The foci of the ellipse are $(\pm a_e e, 0) = (\pm 13 	imes \frac{12}{13}, 0) = (\pm 12, 0)$. Since the hyperbola passes through $(\pm 12, 0)$,we have $\frac{12^2}{a^2} - \frac{0}{b^2} = 1$,which gives $a^2 = 144$. The eccentricity of the hyperbola $e'$ is given by $e' = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{b^2}{144}}$. Given that the product of eccentricities is $1$,we have $e 	imes e' = 1$. $\frac{12}{13} 	imes \sqrt{1 + \frac{b^2}{144}} = 1$. $\sqrt{1 + \frac{b^2}{144}} = \frac{13}{12}$. Squaring both sides,$1 + \frac{b^2}{144} = \frac{169}{144}$. $\frac{b^2}{144} = \frac{169}{144} - 1 = \frac{25}{144}$. Thus,$b^2 = 25$. The equation of the hyperbola is $\frac{x^2}{144} - \frac{y^2}{25} = 1$.

$A$ hyperbola passes through a focus of the ellipse $\frac{x^2}{169}+\frac{y^2}{25}=1$. Its transverse and conjugate axes coincide respectively with the major and minor axes of the ellipse. The product of their eccentricities is $1$. Then,the equation of the hyperbola is

Similar Questions

Tangents are drawn to the hyperbola $x^2 - 9y^2 = 9$ from the point $(3, 2)$. The area of the triangle formed by the tangents and the chord of contact is . . . . . . sq units.

Let $PQ$ be a chord of the hyperbola $\frac{x^2}{4} - \frac{y^2}{b^2} = 1$,perpendicular to the $x$-axis such that $OPQ$ is an equilateral triangle,where $O$ is the centre of the hyperbola. If the eccentricity of the hyperbola is $\sqrt{3}$,then the area of the triangle $OPQ$ is:

Let the product of the focal distances of the point $P(4, 2\sqrt{3})$ on the hyperbola $H: \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ be $32$. Let the length of the conjugate axis of $H$ be $p$ and the length of its latus rectum be $q$. Then $p^2 + q^2$ is equal to ......

The foci of a hyperbola are $(\pm 3, 0)$ and the equation of a tangent is $2x + y - 4 = 0$. Find the equation of the hyperbola.

The radius of the director circle of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module