Question

If the latus rectum of an ellipse $x^2{\tan }^2\phi +y^2{\sec }^2\phi =$ $1$ is $1/2,$ then $\phi$ is

• A. $\pi /2$
• B. $\pi /6$
• C. $\pi /3$
• D. $5$ $\pi /12$

Answer 1

The correct option is D $5$ $\pi /12$

Given $x^{2}ta{n}^2\varphi +y^{2}se{c}^2\varphi =1$

$\to \frac{x^2}{\left(1/ta{n}^2\varphi \right)}+\frac{y^2}{\left(1/se{c}^2\varphi \right)}=1$

$a=\pm \frac{1}{tan\varphi },b=\pm \frac{1}{sec\varphi }$

and $\to e^2=1-\frac{b^2}{a^2}$

$\to e^2=1-\frac{1/se{c}^2\varphi }{1/ta{n}^2\varphi }=1-\frac{ta{n}^2\varphi }{se{c}^2\varphi }$

$\to e^2=1-si{n}^2\varphi =co{s}^2\varphi$

length of latus rectum

$\left(L{L}^{\prime }\right)=\frac{2b^2}{a}=2a(1-e^2)$

$\to 2a(1-co{s}^2\varphi )=2a.si{n}^2\varphi =\frac{1}{2}$ (Given)

$\therefore 2.\frac{cos\varphi }{sin\varphi }si{n}^2\varphi =\frac{1}{2}$

$\to 2cos\varphi sin\varphi =\frac{1}{2}$

$\to si{n}^2\varphi =\frac{1}{2}$

$\to 2\varphi =\frac{\pi }{6},\frac{5\pi }{6}$

$\therefore \varphi =\frac{\pi }{12}$ or

$\varphi =\frac{5\pi }{12}$