Question

Find the relation obtained by eliminating $\theta$ from the equation $x=a\cos \theta +b\sin \theta$ and $y=a\sin \theta -b\cos \theta$

• A. $x^2+y^2=a^2-b^2$
• B. $x^2-y^2=a^2+b^2$
• C. $x^2-y^2=a^2-b^2$
• D. $x^2+y^2=a^2+b^2$

Answer 1

Answer: (D)

$x^2+y^2=a^2+b^2$

Given, $x=a\cos \theta +b\sin \theta$ and $y=a\sin \theta -b\cos \theta$

Squaring them and adding we get

$x^2+y^2=a^2{\cos }^2\theta +b^2{\sin }^2\theta +2ab\cos \theta \sin \theta +a^2{\sin }^2\theta +b^2{\cos }^2\theta -2ab\cos \theta \sin \theta$

$\Rightarrow x^2+y^2=a^2{\cos }^2\theta +b^2{\sin }^2\theta +a^2{\sin }^2\theta +b^2{\cos }^2\theta$

$\Rightarrow x^2+y^2=a^2({\cos }^2\theta +{\sin }^2\theta )+b^2({\sin }^2\theta +{\cos }^2\theta )$

$\Rightarrow x^2+y^2=a^2+b^2$