Question:
If $x=a \sin \theta$ and $y=b \cos \theta$, what is the value of $b^{2} x^{2}+a^{2} y^{2}$ ?
Solution:
Given:
$x=a \sin \theta$ and $y=b \cos \theta$
So,
$b^{2} x^{2}+a^{2} y^{2}=b^{2}(a \sin \theta)^{2}+a^{2}(b \cos \theta)^{2}$
$=a^{2} b^{2} \sin ^{2} \theta+a^{2} b^{2} \cos ^{2} \theta$
$=a^{2} b^{2}\left(\sin ^{2} \theta+\cos ^{2} \theta\right)$
We know that, $\sin ^{2} \theta+\cos ^{2} \theta=1$
Therefore, $b^{2} x^{2}+a^{2} y^{2}=a^{2} b^{2}$
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