For 0

Question

For $0<	heta<\pi / 2$, if the eccentricity of the hyperbola

$x^2-y^2 \operatorname{cosec}^2 	heta=5$ is $\sqrt{7}$ times eccentricity of the

ellipse $x^2 \operatorname{cosec}^2 	heta+y^2=5$, then the value of $	heta$ is :

Accepted Answer

To find the value of $	heta$, we need to determine the relationship between the eccentricities of the given hyperbola and ellipse. Let's start by writing down the standard forms of ellipse and hyperbola and then relate them to the given equations.

The standard form of an ellipse is:

$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

The eccentricity $e$ of an ellipse is given by:

$e = \sqrt{1 - \frac{b^2}{a^2}}\ 	ext{for}\ a > b$

For the given ellipse:

$x^2 \operatorname{cosec}^2 	heta + y^2 = 5$

We can compare it with the standard form by writing it as:

$\frac{x^2}{\frac{5}{\operatorname{cosec}^2 	heta}} + \frac{y^2}{5} = 1$

From this we have $a^2 = \frac{5}{\operatorname{cosec}^2 	heta}$ and $b^2 = 5$. The eccentricity of the ellipse (let's call it $e_1$) is:

$e_1 = \sqrt{1 - \frac{5}{a^2}} = \sqrt{1 - \frac{5}{\frac{5}{\operatorname{cosec}^2 	heta}}} = \sqrt{1 - \operatorname{sin}^2 	heta} = \operatorname{cos} 	heta$

The standard form of a hyperbola is:

$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

The eccentricity $e$ of a hyperbola is given by:

$e = \sqrt{1 + \frac{b^2}{a^2}}$

For the given hyperbola:

$x^2 - y^2 \operatorname{cosec}^2 	heta = 5$

We can rewrite it as:

$\frac{x^2}{5} - \frac{y^2}{\frac{5}{\operatorname{cosec}^2 	heta}} = 1$

From this we have $a^2 = 5$ and $b^2 = \frac{5}{\operatorname{cosec}^2 	heta}$. The eccentricity of the hyperbola (let's call it $e_2$) is:

$e_2 = \sqrt{1 + \frac{\frac{5}{\operatorname{cosec}^2 	heta}}{5}} = \sqrt{1 + \operatorname{sin}^2 	heta}$

According to the question the eccentricity of the hyperbola is $\sqrt{7}$ times the eccentricity of the ellipse, so we can write:

$e_2 = \sqrt{7} \cdot e_1$

Substitute $e_1$ and $e_2$ from the above:

$\sqrt{1 + \operatorname{sin}^2 	heta} = \sqrt{7} \cdot \operatorname{cos} 	heta$

Square both sides to eliminate the square root:

$1 + \operatorname{sin}^2 	heta = 7 \cdot \operatorname{cos}^2 	heta$

$1 + \operatorname{sin}^2 	heta = 7 \cdot (1 - \operatorname{sin}^2 	heta)$

$1 + \operatorname{sin}^2 	heta = 7 - 7 \cdot \operatorname{sin}^2 	heta$

$8 \cdot \operatorname{sin}^2 	heta = 6$

$\operatorname{sin}^2 	heta = \frac{6}{8} = \frac{3}{4}$

$\operatorname{sin} 	heta = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$

Looking at the options provided and knowing the sine values for common angles, we can see that $\operatorname{sin} 	heta = \frac{\sqrt{3}}{2}$ corresponds to $	heta = \frac{\pi}{3}$.

The correct answer is:

Option C.

$\frac{\pi}{3}$

Solutions · 1