Question

If $x=3\sec \theta -2$ and $y=3\tan \theta +2$, then which of the following equation in $x,y$ is correct?

• A. $x^2-y^2+4x+4y=9$
• B. $x^2+y^2-2x+4y=0$
• C. $x^2-y^2-4x+4y=0$
• D. $x^2+y^2+2x-4y=9$

Answer 1

The correct option is A $x^2-y^2+4x+4y=9$

Given:
$x=3\sec \theta -2,y=3\tan \theta +2$
$\Rightarrow \sec \theta =\frac{x+2}{3},\tan \theta =\frac{y-2}{3}$
Using trigonometric identity,
${\sec }^2\theta -{\tan }^2\theta =1$
$\Rightarrow {\left(\frac{x+2}{3}\right)}^2-{\left(\frac{y-2}{3}\right)}^2=1\Rightarrow x^2+4x+4-y^2+4y-4=9$
$\Rightarrow x^2-y^2+4x+4y=9$