Question:
If $x=a \cos \theta$ and $y=b \sin \theta$, then $b^{2} x^{2}+a^{2} y^{2}=$
(a) $a^{2} b^{2}$
(b) $a b$
(c) $a^{4} b^{4}$
(d) $a^{2}+b^{2}$
Solution:
Given:
$x=a \cos \theta, y=b \sin \theta$
So,
$b^{2} x^{2}+a^{2} y^{2}$
$=b^{2}(a \cos \theta)^{2}+a^{2}(b \sin \theta)^{2}$
$=b^{2} a^{2} \cos ^{2} \theta+a^{2} b^{2} \sin ^{2} \theta$
$=b^{2} a^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)$
We know that,
$\sin ^{2} \theta+\cos ^{2} \theta=1$
Therefore, $b^{2} x^{2}+a^{2} y^{2}=a^{2} b^{2}$
Hence, the correct option is (a).