Question

If $\frac{x}{a}\cos \theta +\frac{y}{b}\sin \theta =1and\frac{x}{a}\sin \theta -\frac{y}{b}\cos \theta =1$, prove that $\frac{x^2}{a^2}+\frac{y^2}{b^2}=2$

Answer 1

We have,

$\frac{x}{a}\cos \theta +\frac{y}{b}\sin \theta =1$

On squaring both sides, we get

$\frac{x^2}{a^2}{\cos }^2\theta +\frac{y^2}{b^2}{\sin }^2\theta +\frac{2xy}{ab}\sin \theta \cos \theta =1$ $.........\left(1\right)$

Simliarly,

$\frac{x}{a}\sin \theta -\frac{y}{b}\cos \theta =1$

$\frac{x^2}{a^2}{\sin }^2\theta +\frac{y^2}{b^2}{\cos }^2\theta -\frac{2xy}{ab}\sin \theta \cos \theta =1$ $.........\left(2\right)$

On adding both equations, we get

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

Since,

${\cos }^2\theta +{\sin }^2\theta =1$

Therefore,

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

Hence, this is the answer.