Question

x/a cos theta + y/b sin theta =1 find the value of x^2/a^2 + y^2/b^2​

Answer 1

Correct Question :

If x/a cosθ +y/b sinθ =1 and x/a sinθ- y/bcosθ=1 , then prove that x²/a²+y² /b² =2

Given :

$\sf \dfrac{xcos\theta}{a}+\dfrac{ysin\theta}{b}=1...(1)\\\\\sf \dfrac{xsin\theta}{a}-\dfrac{ycos\theta}{b}=1...(2)$

To Find :

$\sf Value\ of\ \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}$

Solution :

Squaring (1) on both sides ,

$\\ \sf \bigg(\dfrac{xcos\theta}{a}+\dfrac{ysin\theta}{b}\bigg)^2=(1)^2$

$\\ \to \sf \dfrac{x^2cos^2\theta}{a^2}+\dfrac{y^2sin^2\theta}{b^2}+\dfrac{2xysin \theta cos \theta}{ab}=1...(3)$

Squaring (2) on both sides ,

$\\ \sf \bigg(\dfrac{xsin\theta}{a}-\dfrac{ycos\theta}{b}\bigg)^2=(1)^2$

$\\ \to \sf \dfrac{x^2sin^2\theta}{a^2}+\dfrac{y^2cos^2\theta}{b^2}-\dfrac{2xysin\theta cos\theta}{ab}=1...(4)$

Add (3) and (4) ,

$\\ \to \sf \dfrac{x^2cos^2\theta}{a^2}+\dfrac{y^2sin^2\theta}{b^2}+\dfrac{2xysin \theta cos \theta}{ab}+\dfrac{x^2sin^2\theta}{a^2}+\dfrac{y^2cos^2\theta}{b^2}-\dfrac{2xysin \theta cos \theta}{ab}=1+1$

$\\ \to \sf \dfrac{x^2}{a^2}(sin^2\theta+cos^2\theta)+\dfrac{y^2}{b^2}(sin^2\theta+cos^2\theta)=2$

Remember Trigonometric Identity ,

→   sin²θ + cos²θ = 1   ←

$\\ \to \sf \dfrac{x^2}{a^2}(1)+\dfrac{y^2}{b^2}(1)=2$

$\\ \sf \pink{\leadsto \dfrac{x^2}{a^2}+\dfrac{x^2}{b^2}=2}\ \; \bigstar$