Let the eccentricity of the hyperbola frac{x^ | vedclass

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

Question

(A) The given equation of the ellipse is $x^{2}+9y^{2}=9$,which can be written as $\frac{x^{2}}{9}+\frac{y^{2}}{1}=1$. Here,$a^{2}=9$ and $b^{2}=1$,so

Vedclass · Accepted Answer

$8:1$

Vedclass · Answer

(A) The given equation of the ellipse is $x^{2}+9y^{2}=9$,which can be written as $\frac{x^{2}}{9}+\frac{y^{2}}{1}=1$. Here,$a^{2}=9$ and $b^{2}=1$,so $a=3$ and $b=1$. The eccentricity $e_{e}$ of the ellipse is given by $e_{e} = \sqrt{1-\frac{b^{2}}{a^{2}}} = \sqrt{1-\frac{1}{9}} = \sqrt{\frac{8}{9}} = \frac{\sqrt{8}}{3}$. The eccentricity $e_{h}$ of the hyperbola is the reciprocal of $e_{e}$,so $e_{h} = \frac{3}{\sqrt{8}}$. For the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$,the eccentricity is $e_{h} = \sqrt{1+\frac{b^{2}}{a^{2}}}$. Squaring both sides,we get $e_{h}^{2} = 1+\frac{b^{2}}{a^{2}} = \frac{9}{8}$. Thus,$\frac{b^{2}}{a^{2}} = \frac{9}{8}-1 = \frac{1}{8}$. Therefore,the ratio $a^{2}:b^{2} = 8:1$.

Let the eccentricity of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ be reciprocal to that of the ellipse $x^{2}+9y^{2}=9$. Then the ratio $a^{2}:b^{2}$ equals:

Similar Questions

The auxiliary circle of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is given by:

The locus of the midpoints of the chord of the circle $x^{2}+y^{2}=25$ which is tangent to the hyperbola $\frac{x^{2}}{9}-\frac{y^{2}}{16}=1$ is

If $x=9$ is a chord of contact of the hyperbola $x^2-y^2=9$,then the equation of the tangent at one of the points of contact is

The equation of the tangent to the hyperbola $4y^2 = x^2 - 1$ at the point $(1, 0)$ is

The equation of the hyperbola having its eccentricity $e = 2$ and the distance between its foci as $8$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module