Question

If $x=3(\cos t+\sin t);y=2(\cos t-\sin t)$ represents a conic, then its foci are

• A. $(\pm \sqrt{10},0)$
• B. $(\pm \sqrt{13},0)$
• C. $(0,\pm \sqrt{13})$
• D. $(0,\pm \sqrt{10})$

Answer 1

The correct option is A $(\pm \sqrt{10},0)$

$x=3(\cos t+\sin t);y=2(\cos t-\sin t)$
$\Rightarrow \frac{x}{3}=\cos t+\sin t\cdots \left(1\right)$
$\Rightarrow \frac{y}{2}=\cos t-\sin t\cdots \left(2\right)$

Now, squaring and adding equation $\left(1\right)$ and $\left(2\right),$ we get
$\frac{x^2}{9}+\frac{y^2}{4}=2({\sin }^{2}t+{\cos }^{2}t)$
$\Rightarrow \frac{x^2}{18}+\frac{y^2}{8}=1$
$\Rightarrow a^2=18,b^2=8(a>b)$
$\Rightarrow e=\sqrt{1-\frac{b^2}{a^2}}$
$\Rightarrow e=\sqrt{1-\frac{8}{18}}$
$\Rightarrow e=\frac{\sqrt{10}}{\sqrt{18}}$
$\therefore$ foci are $(\pm \sqrt{10},0)$