Q. A Basic Terms of Conics is defined by the equations $x = - 1 + \sec{t,\ y = 2 + 3\tan{t.}}$ The coordinates of the foci are
A) $( - 1 - \sqrt{10},2)$ and $( - 1 + \sqrt{10},2)$
B) $( - 1 - \sqrt{8},2)$ and $( - 1 + \sqrt{8},2)$
C) $( - 1,2 - \sqrt{8})$ and $( - 1,\ 2 + \sqrt{8})$
D) $( - 1,\ 2 - \sqrt{10})$ and $( - 1,\ 2 + \sqrt{10})$
Explanation
(a)
Given, $x + 1 = \sec t,\ \ \ \frac{y - 2}{3} = \tan\ t$
Since $\sec^{2}{t - \tan^{2}{= 1}}$
$\therefore\ \frac{(x + 1)^{2}}{1} - \frac{(y - {2)}^{2}}{9} = 1$
Now, $e = \sqrt{1 + \frac{b^{2}}{a^{2}}} = \sqrt{1 + \frac{9}{1}} = \sqrt{10}$
$\therefore$ Foci $= ( - 1 \pm ae,2)$
$= \ \left( - 1 - \sqrt{10},\ 2 \right)$ and $\left( - 1 + \sqrt{10},2 \right)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.