Question

question no. 15 (1st part )

Answer 1

(x/a)cos(θ) + (y/b)sin(θ) = 1 and (x/a)sin(θ) - (y/b)cos(θ) = 1

Squaring both equations and adding them:

(x²/a²)*cos²(θ) + 2(xy/ab)cos(θ)sin(θ) + (y²/b²)*sin²(θ) + (x²/a²)sin²(θ) - 2(xy/ab)cos(θ)sin(θ) + (y²/b²)*cos²(θ) = 2

Therefore:

x²/a² + y²/b² = 2