Mathematics - Ellipse Question with Solution | TestHub

The given ellipse is $$\begin{array}{rl}& \frac{x^2}{4}+\frac{y^2}{3}=1\Rightarrow \alpha =2,\beta =\sqrt{3}\\ & \therefore 3=4\left(1-e^2\right)\Rightarrow e=\frac{1}{2}\\ & \end{array}$$ $$\therefore ae=1$$ Hence the eccentricity $e_1$, of the hyperbola is given by $$\begin{array}{rlrl}& & 1& =e_1\sin \theta \\ \Rightarrow & & e_1& =\mathrm{cosec}\theta \\ \Rightarrow & b^2& ={\sin }^2\theta \left({\mathrm{cosec}}^2\theta -1\right)={\cos }^2\theta & \end{array}$$ Hence, the hyperbola is $$\frac{x^2}{{\sin }^2\theta }-\frac{y^2}{{\cos }^2\theta }=1\text{ or }{x}^2{\mathrm{cosec}}^2\theta -y^2{\sec }^2\theta =1$$