Question

If $x=2(\cos t-\sin t)$; $y=3(cost+sint)$ represents a conic, 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 D $(0,\pm \sqrt{10})$

$x=2(\cos t-\sin t)$

$y=3(\cos t+\sin t)$

${\left(\frac{x}{2}\right)}^2+{\left(\frac{y}{3}\right)}^2=(\cos t-\sin t{)}^2+(\cos t+\sin t{)}^2=2({\cos }^{2}t+{\sin }^{2}t)$

$\frac{x^2}{4}+\frac{y^2}{9}=2$

$\frac{x^2}{8}+\frac{y^2}{18}=1$

Eccentricity is $e^2=\frac{b^2-a^2}{b^2}=\frac{18-8}{18}=\frac{5}{9}$

$e=\frac{\sqrt{5}}{3}$

Foci are $(0,\pm be)=(0,\pm \sqrt{18}\times \frac{\sqrt{5}}{3})=(0,\pm \sqrt{10})$