Question

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

Answer 1

Given,

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

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

Squaring and adding (1) and (2), we get,

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

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

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

Hence proved