Question

A conic section is defined by the equations $ x=-1+\sec t,\text{ }y=2+3\text{ }tan\text{ }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}) $
• 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) $