TRIGNOMETRY HELP ..PLEAAAAASEEEEEE? WILL BE MOST GREATFUL!?

Question

iF(x/a)cosθ = (y/b)sinθ
AND ax/cosθ - by/sinθ = a² - b²
PROVE THAT x²/a² + y²/b² = 1?

Answer 1

The given equations are (x/a)cosθ = (y/b)sinθ   --------------(1)

   ax/cosθ - by/sinθ = a² - b²                                    --------------(2)

(x/a)cosθ = (y/b)sinθ  = k

x = k(a)cosθ and y = k(b)sinθ

Put the values x and y in the equation (2)

a(k(a)cosθ)/cosθ - b(k(b)sinθ)/sinθ = a² - b²

Cancel common terms.

ka² - kb² = a² - b²

k(a² - b²) = a² - b²

k = (a² - b²)/(a² - b²)

k = 1

Now x²/a² + y²/b² = 1

= (k(a)cosθ)² /a² + (k(b)sinθ)²/b²

= ( k²a²cos²θ)/a² + (k²b²sin² θ)/b²

Cancel common terms.

= k²(cos²θ + sin² θ)

= k²(1)

= 1