Let one focus of the hyperbola H : frac{x^2}{ | vedclass

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

Question

(C) Given the focus $ae = \sqrt{10}$ and the directrix $\frac{a}{e} = \frac{9}{\sqrt{10}}$.Multiplying these,we get $a^2 = \sqrt{10} 	imes \frac{9}{\

Vedclass · Accepted Answer

$16$

Vedclass · Answer

(C) Given the focus $ae = \sqrt{10}$ and the directrix $\frac{a}{e} = \frac{9}{\sqrt{10}}$.Multiplying these,we get $a^2 = \sqrt{10} 	imes \frac{9}{\sqrt{10}} = 9$,so $a = 3$.Then $e = \frac{\sqrt{10}}{a} = \frac{\sqrt{10}}{3}$.For a hyperbola,$b^2 = a^2(e^2 - 1) = a^2e^2 - a^2 = (ae)^2 - a^2$.Substituting the values,$b^2 = 10 - 9 = 1$.The length of the latus rectum $l = \frac{2b^2}{a} = \frac{2(1)}{3} = \frac{2}{3}$.Now,$9(e^2 + l) = 9\left(\left(\frac{\sqrt{10}}{3}ight)^2 + \frac{2}{3}ight) = 9\left(\frac{10}{9} + \frac{2}{3}ight) = 10 + 6 = 16$.

Let one focus of the hyperbola $H : \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ be at $(\sqrt{10}, 0)$ and the corresponding directrix be $x = \frac{9}{\sqrt{10}}$. If $e$ and $l$ respectively are the eccentricity and the length of the latus rectum of $H$,then $9(e^2 + l)$ is equal to:

Similar Questions

The equation $\frac{1}{r} = \frac{1}{8} + \frac{3}{8} \cos \theta$ represents:

If $e$ and $e'$ are the eccentricities of a hyperbola and its conjugate hyperbola respectively,then:

The locus of the point of intersection of tangents at the ends of a normal chord of the hyperbola $x^2 - y^2 = a^2$ is

If $x = 9$ is the chord of contact of the hyperbola $x^2 - y^2 = 9$,then the equation of the corresponding pair of tangents is

Let the hyperbola be $x^2 - 2y^2 - 2\sqrt{2}x - 4\sqrt{2}y - 6 = 0$. Let $A$ be a vertex of the hyperbola. Let $B$ be an endpoint of the latus rectum near $A$. If $C$ is the focus of the hyperbola nearest to $A$,find the area of triangle $ABC$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module