If x = a cos alpha and y = b sin alpha. Find the value of b^2x^2 + a^2y^2..... someone help

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If x = a cos alpha and y = b sin alpha. Find the value of b^2x^2 + a^2y^2..... someone help

Mathematics
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If x = a cos alpha and y = b sin alpha Then a=x/cos alpha and b= y/sin alpha Then b^2x^2 + a^2y^2 is: \(\Large (\frac{y}{sin \alpha})^2(a* cos \alpha )^2 + (\frac{x}{cos \alpha})^2(b* sin \alpha)^2\)
consider \(bx=ab\cos\alpha\) and \(ay=ab\sin\alpha\), so \(b^2x^2+a^2y^2=(bx)^2+(ay)^2=ab^2(\cos^2\alpha+\sin^2\alpha)=a^2b^2\)
oops, the third part of hte second line should read \(a^2b^2(\cos^2\alpha+\sin^2\alpha)\)

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Other answers:

@oldrin.bataku and @Zale101 the answer is a^2b^2 i just want to knw the method can uh explain briefly pls :D
$$b^2x^2+a^2y^2=b^2a^2\cos^2\alpha+a^2b^2\sin^2\alpha=a^2b^2(\cos^2\alpha+\sin^2\alpha)$$ by plugging in, expanding, and factoring, yes?
then the Pythagorean identity tells us \(\cos^2\alpha+\sin^2\alpha=1\), so that simplifies to \(a^2b^2\cdot1=a^2b^2\)
ok Thank uh @oldrin.bataku :)

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