Question

The locus of the point $x=a(\cos \theta +\sin \theta )$ , $y=b(\cos \theta -\sin \theta )$ is

• A. $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
• B. $\frac{x^2}{a^2}+\frac{y^2}{b^2}=2$
• C. $\frac{x^2}{a^2}+\frac{y^2}{b^2}=\frac{1}{2}$
• D. $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

Answer 1

Answer: (B)

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=2$

It is given that

$x=a(\cos \theta +\sin \theta )$

$y=b(\cos \theta -\sin \theta )$

Now $bx=ab(\cos \theta +\sin \theta )$ ...(i)

$ay=ab(\cos \theta -\sin \theta )$ ...(ii)

Hence, $b^2{x}^2+a^2{y}^2$

$=a^2{b}^2({\cos }^2\theta +{\sin }^2\theta +2\cos \theta \sin \theta +{\cos }^2\theta +{\sin }^2\theta -2\cos \theta \sin \theta )$

$=a^2{b}^2\left(2\right({\cos }^2\theta +{\sin }^2\theta \left)\right)$

$=2a^2{b}^2$

Hence, $b^2{x}^2+a^2{y}^2=2a^2{b}^2$

Dividing the entire equation by $a^2{b}^2$ gives us

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=2$