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

1. Zale101

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: $$\Large (\frac{y}{sin \alpha})^2(a* cos \alpha )^2 + (\frac{x}{cos \alpha})^2(b* sin \alpha)^2$$

2. anonymous

consider $$bx=ab\cos\alpha$$ and $$ay=ab\sin\alpha$$, so $$b^2x^2+a^2y^2=(bx)^2+(ay)^2=ab^2(\cos^2\alpha+\sin^2\alpha)=a^2b^2$$

3. anonymous

oops, the third part of hte second line should read $$a^2b^2(\cos^2\alpha+\sin^2\alpha)$$

4. anonymous

@oldrin.bataku and @Zale101 the answer is a^2b^2 i just want to knw the method can uh explain briefly pls :D

5. anonymous

$$b^2x^2+a^2y^2=b^2a^2\cos^2\alpha+a^2b^2\sin^2\alpha=a^2b^2(\cos^2\alpha+\sin^2\alpha)$$ by plugging in, expanding, and factoring, yes?

6. anonymous

then the Pythagorean identity tells us $$\cos^2\alpha+\sin^2\alpha=1$$, so that simplifies to $$a^2b^2\cdot1=a^2b^2$$

7. anonymous

ok Thank uh @oldrin.bataku :)