Question

The locus of the point $tan\theta +sin\theta ,tan\theta -sin\theta$ is

• A. ${\left(x^2{y}^2\right)}^{2/3}+{\left(x{y}^2\right)}^{2/3}=1$
• B. $x^2-y^2=xy$
• C. $x^2-y^2=12xy$
• D. ${(x}^2-y^2{)}^2=16xy$

Answer 1

The correct option is D ${(x}^2-y^2{)}^2=16xy$

We have,

$x=\tan \theta +\sin \theta .......\left(1\right)$

$y=\tan \theta -\sin \theta .......\left(2\right)$

Can it take all allowed real value it to $[-\pi ,\pi ]to[\frac{-\pi }{2},\frac{\pi }{2}]$

On adding and subtracting equation (1) and (2) and we get,

$x+y=2\tan \theta$

$x-y=2\sin \theta$

On divide and we get,

$\frac{x+y}{x-y}=\frac{\tan \theta }{\sin \theta }$

$\frac{x+y}{x-y}=\sec \theta$

On squaring both side and we get,

${\left(\frac{x+y}{x-y}\right)}^2={\sec }^2\theta$

If $x+y=2\tan \theta$

On squaring and we get,

${(x+y)}^2=4{\tan }^2\theta$

$\Rightarrow {(x+y)}^2=4({\sec }^2\theta -1)$

$\Rightarrow {(x+y)}^2=4(\frac{{(x+y)}^2}{{(x-y)}^2}-1)$

$\Rightarrow {(x+y)}^2=4\left(\frac{{(x+y)}^2-{(x-y)}^2}{{(x-y)}^2}\right)$

$\Rightarrow {(x+y)}^2{(x-y)}^2=4({(x+y)}^2-{(x-y)}^2)$

$\Rightarrow {(x^2-y^2)}^2=4\left[4xy\right]$

$\Rightarrow {(x^2-y^2)}^2=16xy$

Hence, this is the answer.