Question

x/acosθ+y/bsinθ=1 and x/asinθ-y/bcosθ=1, prove that x²/a²+y²/b²=2.

Answer 1

$\text{Given equations are}$

$\displaystyle\frac{x}{a}cos\theta+\frac{y}{b}sin\theta=1$ .......(1)

$\displaystyle\frac{x}{a}sin\theta-\frac{y}{b}cos\theta=1$ .......(2)

$\text{Squaring and adding (1) and (2)}$

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

$\displaystyle\frac{x^2}{a^2}cos^2\theta+\frac{y^2}{b^2}sin^2\theta+2\,\frac{xy}{ab}cos\theta\,sin\theta+\frac{x^2}{a^2}sin^2\theta+\frac{y^2}{b^2}cos^2\theta-2\,\frac{xy}{ab}cos\theta\,sin\theta=2$

$\displaystyle\frac{x^2}{a^2}cos^2\theta+\frac{y^2}{b^2}sin^2\theta+\frac{x^2}{a^2}sin^2\theta+\frac{y^2}{b^2}cos^2\theta=2$

$\displaystyle\frac{x^2}{a^2}(cos^2\theta+sin^2\theta)+\frac{y^2}{b^2}(sin^2\theta+cos^2\theta)=2$

$\displaystyle\frac{x^2}{a^2}(1)+\frac{y^2}{b^2}(1)=2$

$\implies\boxed{\bf\frac{x^2}{a^2}+\frac{y^2}{b^2}=2}$