x/a cosθ + y/b sinθ = 1 and x/a sinθ - y/bcosθ = 1, prove that x^2/a^2+y^2/b^2 = 2

Question

\(\frac{x}{a}\) cosθ + \(\frac{y}{b}\) sinθ =1 and \(\frac{x}{a}\) sinθ - \(\frac{y}{b}\) cosθ = 1, prove that \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) = 2.

Answer 1

\(\Big(\frac{x}{a}cosθ+\frac{y}{b}sinθ\Big)^2\) + \(\Big(\frac{x}{a}sinθ-\frac{y}{b}cosθ\Big)^2\)=2

\(\frac{x^2}{a^2}cos^2θ+\frac{y^2}{b^2}sin^2θ\) + \(2\frac{xy}{ab}cosθsinθ+\frac{x^2}{a^2}sin^2θ\) + \(\frac{y^2}{b^2}cos^2θ-2\frac{xy}{ab}sinθcosθ=2\) 

Hence Proved.

Answer 2

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