Question

Find the locus of the curve represented by $x=sec\theta +1andy=tan\theta -1$, where $\theta$ is a variable.

• A. (x-1)² + (y+1)² = 1
• B. (x-1)² - (y+1)² = 1
• C. (x+1)² - (y-1)² = 0
• D. (x+1)² + (y-1)² = 1

Answer 1

The correct option is B

(x-1)² - (y+1)² = 1

Locus is the collection of points satisfying some given condition. We use the given condition to find the equation of the curve, usually after eliminating variables given in the condition.

In this case $\theta$ is the variable given, we have $x=sec\theta +1andy=tan\theta -1$. We want to eliminate $\theta$.

After seeing the above relations and going through the options we can guess that we will use the identity $se{c}^2\theta -ta{n}^2\theta =1$. For that, we will find $sec\theta andtan\theta$ in terms of x and y.

$\Rightarrow sec\theta =x-1andtan\theta =y+1$
$se{c}^2\theta -ta{n}^2\theta =1\Rightarrow (x-1{)}^2-(y+1{)}^2=1$
This is the locus of the points.