Question

For some$\theta \in (0,\frac{\mathrm{\pi }}{2})$, if the eccentricity of the hyperbola, $x^2-y^2{\sec }^2\left(\theta \right)=10$ is $\sqrt{5}$ times the eccentricity of the ellipse, $x^2{\sec }^2\left(\theta \right)+y^2=5$, then the length of the latus rectum of the ellipse, is:

• A. $\frac{4\sqrt{5}}{3}$
• B. $\frac{2\sqrt{5}}{3}$
• C. $2\sqrt{6}$
• D. $\sqrt{30}$

Answer 1

The correct option is A

$\frac{4\sqrt{5}}{3}$

The explanation for the correct option:

Step1.Find the value ${\mathbf{a}}^{\mathbf{2}}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{b}}^{\mathbf{2}}\mathbf{}$ of hyperbola and ellipse:

Given, For,$\mathrm{\theta }\in (0,\frac{\mathrm{\pi }}{2})$

The given equation of hyperbola is $x^2-y^2{\sec }^2\left(\theta \right)=10$

We know that

The general equation of the hyperbola is $\begin{array}{rcl}\frac{{\mathbf{x}}^{\mathbf{2}}}{{\mathbf{a}}^{\mathbf{2}}}\mathbf{-}\frac{{\mathbf{y}}^{\mathbf{2}}}{{\mathbf{b}}^{\mathbf{2}}}& \mathbf{=}& \mathbf{1}\end{array}$

Now, we have $x^2-y^2{\sec }^2\left(\theta \right)=10$

$\Rightarrow$ $\begin{array}{rcl}\frac{x^2}{10}-\frac{y^2}{10{\cos }^2\left(\theta \right)}& =& 1\end{array}$

$\begin{array}{rcl}\therefore a^2& =& 10,b^2=10{\cos }^2\left(\theta \right)\end{array}$

The given equation of an ellipse is $x^2{\sec }^2\left(\theta \right)+y^2=5$

We know that

The general equation of the ellipse is $\begin{array}{rcl}\frac{{\mathbf{x}}^{\mathbf{2}}}{{\mathbf{a}}^{\mathbf{2}}}\mathbf{+}\frac{{\mathbf{y}}^{\mathbf{2}}}{{\mathbf{b}}^{\mathbf{2}}}& \mathbf{=}& \mathbf{1}\end{array}$

Now, we have $x^2{\sec }^2\left(\theta \right)+y^2=5$

$\Rightarrow$ $\begin{array}{rcl}\frac{x^2}{5{\cos }^2\left(\theta \right)}+\frac{y^2}{5{\cos }^2\left(\theta \right)}& =& 1\end{array}$

$\begin{array}{rcl}\therefore a^2& =& 5,b^2=5{\cos }^2\left(\theta \right)\end{array}$

Step2. Calculate the value of $\cos \left(\theta \right)$:

The eccentricity of the hyperbola is

$\mathbf{\text{'}}{\mathbf{e}}_{\mathbf{1}}\mathbf{\text{'}}\mathbf{}\mathbf{=}\sqrt{\mathbf{1}\mathbf{+}\frac{{\mathbf{b}}^{\mathbf{2}}}{{\mathbf{a}}^{\mathbf{2}}}}$

$\Rightarrow$ $e_1=\sqrt{1+\frac{10{\cos }^2\left(\theta \right)}{10}}$

The eccentricity of the ellipse:

$\mathbf{\text{'}}{\mathbf{e}}_{\mathbf{2}}\mathbf{\text{'}}\mathbf{}\mathbf{=}\sqrt{\mathbf{1}\mathbf{-}\frac{{\mathbf{b}}^{\mathbf{2}}}{{\mathbf{a}}^{\mathbf{2}}}}$

$\Rightarrow$ $e_2=\sqrt{1-\frac{5{\cos }^2\left(\theta \right)}{5}}$

As per the given condition, $e_1=\sqrt{5}{e}_2$

Putting the values

$$\begin{gathered} \Rightarrow \sqrt{1+\frac{10{\cos }^2\left(\theta \right)}{10}}=\left(\sqrt{5}\right)\sqrt{1-\frac{5{\cos }^2\left(\theta \right)}{5}} \\ \Rightarrow 1+\frac{10{\cos }^2\left(\theta \right)}{10}=5\left(1-\frac{5{\cos }^2\left(\theta \right)}{5}\right) \\ \Rightarrow 1+{\cos }^2\left(\theta \right)=5\left(1-{\cos }^2\left(\theta \right)\right) \\ \Rightarrow 6{\cos }^2\left(\theta \right)=4 \\ \Rightarrow {\cos }^2\left(\theta \right)=\frac{4}{6} \\ \Rightarrow \cos \left(\theta \right)=\frac{2}{3}\left[\mathbf{\because }\mathbf{0}\mathbf{<}\mathit{\theta }\mathbf{<}\frac{\mathbf{\pi }}{\mathbf{2}}\right] \end{gathered}$$

Step3. Calculate the length of the latus rectum of the ellipse:

The length of the latus rectum of the ellipse is $\frac{\mathbf{2}{\mathbf{b}}^{\mathbf{2}}}{\mathbf{a}}$

Now, put the values of $a,b$ we get

$$\begin{gathered} =\frac{2\left(5{\cos }^2\left(\theta \right)\right)}{\sqrt{5}} \\ \mathbf{=}\mathbf{}\frac{2\times 5\times \left({\displaystyle \frac{4}{6}}\right)}{\sqrt{5}} \\ \mathbf{=}\frac{\mathbf{4}\sqrt{\mathbf{5}}}{\mathbf{3}} \end{gathered}$$

Hence, Option(A) is the correct answer.