if x = a cos theta and y = b sin theta , then find b2x2 + a2y2
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HLO FRND!!!!
Here's the answer you are looking for
x = acosθ
y = bsinθ
So,
b²x² + a²y² - a²b²
= b² (acosθ)² + a² (bsinθ)² - a²b²
= b²a²cos²θ + a²b²sin²θ - a²b²
= a²b²cos²θ + a²b²sin²θ - a²b²
= a²b² ( cos²θ + sin²θ - 1)
= a²b² (1 - 1) [cos²θ + sin²θ = 1]
= a²b² (0)
= 0
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