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 $b^{2}=1$,

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 locus of a variable point whose chord of contact with respect to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ subtends a right angle at the origin is

The equation of the normal to the hyperbola $\frac{x^2}{16} - \frac{y^2}{9} = 1$ at the point $(8, 3\sqrt{3})$ is

If the circle $x^{2}+y^{2}=a^{2}$ intersects the hyperbola $xy=c^{2}$ in four points $P(x_{1}, y_{1}), Q(x_{2}, y_{2}), R(x_{3}, y_{3})$ and $S(x_{4}, y_{4})$,then

The $X$ and $Y$ intercepts of the tangent to the hyperbola $\frac{x^2}{20}-\frac{y^2}{5}=1$ which is perpendicular to the line $4x+3y=7$,are respectively

The foci of the hyperbola $4x^2 - 9y^2 - 36 = 0$ are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module