Question

A hyperbola, having the transverse axis of length $2\sin \theta$, is confocal with the ellipse $3x^2+y^2=12$ then, its equation is

• A. $x^2\cos e{c}^2\theta –y^{2}se{c}^2\theta =1$
• B. $x^{2}se{c}^2\theta –y^2\cos e{c}^2\theta =1$
• C. $x^2{\sin }^2\theta –y^2{\cos }^2\theta =1$
• D. $x^2{\cos }^2\theta –y^2{\sin }^2\theta =1$

Answer 1

The correct option is A

$x^2\cos e{c}^2\theta –y^{2}se{c}^2\theta =1$

The explanation for the correct option:

Step 1: Expressing the given information:

The given ellipse is

$\begin{array}{rcl}3x^2+y^2& =& 12\\ & \Rightarrow & \frac{x^2}{4}+\frac{y^2}{3}=1\end{array}$

Hence,

$$\begin{gathered} a^2=4b^2=3 \\ a=2b=\sqrt{3} \end{gathered}$$

Step 2: Calculating the variables for the equation

$\begin{array}{rcl}e& =& \sqrt{1-\frac{3}{4}}\\ & =& \frac{1}{2}\end{array}$

The focus of the ellipse = $(\pm ae,0)=(\pm 1,0)$

Given required hyperbola is confocal to the ellipse

Let, $a\text{'},b\text{'},e\text{'}$ be the transverse axis, conjugate axis, and eccentricity of the hyperbola

$\begin{array}{rcl}a\text{'}e\text{'}& =& 1\\ \sin \theta .e\text{'}& =& 1\\ e\text{'}& =& \frac{1}{\sin \theta }\\ & & \end{array}$

$\begin{array}{rcl}b{\text{'}}^2& =& a{\text{'}}^2(e^2-1)\\ b{\text{'}}^2& =& 1-{\sin }^2\theta \\ & =& {\cos }^2\theta \end{array}$

Hence the required hyperbola is

$\frac{x^2}{{\sin }^2\theta }-\frac{y^2}{{\cos }^2\theta }=1$

Hence, the equation of the hyperbola is $x^2\cos e{c}^2\theta -y^{2}se{c}^2\theta =1$ so, option (A) is the correct option.