Question

A conic section is defined by the equations $x=-1+\sec t,y=2+3tant$ . 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 $(-1,2+\sqrt{10})$
• E. $(\sqrt{10},0)$ and $(-\sqrt{10},0)$

Answer 1

Answer: (A)

Given equations can be rewritten as
$x+1=\sec t,\frac{y-2}{3}=\tan t$
Since, ${\sec }^{2}t-{\tan }^{2}t=1$
$\therefore$ $\frac{{(x+1)}^2}{1}-\frac{{(y-2)}^2}{9}=1$
Now, $e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\frac{9}{1}}=\sqrt{10}$
$\therefore$ $Foci=(-1\pm ae,2)$
$=(-1-\sqrt{10},2)$ and $(-1+\sqrt{10},2)$