Question

$Iftheeccentricityofhyperbola{x}^2-y^{2}se{c}^2\alpha =5is\sqrt{3}timestheeccentricityoftheellipse{x}^{2}se{c}^2\alpha +y^2=25,then\alpha =$

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{2}$

Answer 1

The correct option is B

$\frac{\pi }{4}$

$Thehyperbola{x}^2-y^{2}se{c}^2\alpha =5canberewritteninthefollowingway:$

$\frac{x^2}{5}-\frac{y^2}{5co{s}^2\alpha }=1$

$\text{This is the standard form of a hyperbola,where}{a}^2=5and{b}^2=5co{s}^2\alpha$

$\Rightarrow b^2=a^2(e_2^2-1)$

$\Rightarrow 5co{s}^2\alpha =5(e_2^2-1)$

$\Rightarrow e_1^2=co{s}^2\alpha +1...\left(1\right)$

$Theellipse{x}^{2}se{c}^2\alpha +y^2=25canberewritteninthefollowingway:$

$\frac{x^2}{25co{s}^2\alpha }+\frac{y^2}{25}=1$

$\text{The is the standard form of an ellipse,where}{\alpha }^2=25and{b}^2=25co{s}^2\alpha$

$b^2=a^2(1-e_2^2)$

$\Rightarrow e_2^2=1-co{s}^2\alpha$

$\Rightarrow e_2^2=si{n}^2\alpha$

$Accordingtothequestion,$

$co{s}^2\alpha +1=3\left(si{n}^2\alpha \right)$

$\Rightarrow 2=4si{n}^2\alpha$

$\Rightarrow \alpha =\frac{\pi }{4}$