Question

8th maths s. Chand chapter20

Answer 1

Answer:

if u mean this then

Step-by-step explanation

\bold{\frac{x}{a}cos\theta+\frac{y}{b}sin\theta=1-----(1)}\\\\\bold{\frac{x}{a}sin\theta-\frac{y}{b}cos\theta=1-----(2)}

Squaring both the equations and then adding ,  

\bold{[\frac{x}{a}cos\theta+\frac{y}{b}sin\theta]^2+[\frac{x}{a}sin\theta-\frac{y}{b}cos\theta]^2=1^2+1^2}  

x²/a² cos²θ + y²/b² sin²θ + 2xy/ab sinθ.cosθ + x²/a² sin²θ + y²/b² cos²θ - 2xy/ab sinθ.cosθ = 2  

⇒x²/a² (cos²θ + sin²θ) + y²/b² (sin²θ + cos²θ ) = 2  

⇒x²/a² × 1 + y²/b² × 1 = 2 [ ∵ sin²x + cos²x = 1 from trigonometric identities ]  

∴ x²/a² + y²/b² = 2 , hence proved