Question

The parametric form the curve
$\frac{(x+1{)}^2}{16}-\frac{(y-2{)}^2}{4}=1$ is

• A. $(4\sec \theta ,2\tan \theta )$
• B. $(4\sec \theta +1,2\tan \theta -2)$
• C. $(4\sec \theta -1,2\tan \theta +2)$
• D. $(4\sec \theta -4,2\tan \theta -2)$

Answer 1

The correct option is C

$(4\sec \theta -1,2\tan \theta +2)$

While finding parametric form of the standard hyperbola we effectively equated
$\frac{x}{a}\text{to}\sec \theta \text{and}\frac{y}{b}\text{to}\tan \theta$
so that it satisfies the identity ${\sec }^2\theta -{\tan }^2\theta =1$.
This is true for hyperbola of the form

${\left[\frac{x-h}{a}\right]}^2-{\left[\frac{y-k}{b}\right]}^2=1$
as well
Here
$\frac{x-h}{a}=\sec \theta \text{and}\frac{y-k}{b}=\tan \theta$

$\therefore \frac{x+1}{4}=\sec \theta \text{and}\frac{y-2}{2}=\tan \theta$

$\therefore \text{Parametric form}=(4\sec \theta -1,2\tan \theta +2)$