Question

If $\frac{x}{a}cos\theta +\frac{y}{b}sin\theta =1,\frac{x}{a}sin\theta -\frac{y}{b}cos\theta =1$, then ___.

• A. $x^2+y^2=a^2+b^2$
• B. $\frac{x^2}{a^2}+\frac{y^2}{b^2}=2$
• C. $a^2{x}^2+b^2{y}^2=1$
• D. $x^2-y^2=a^2-b^2$

Answer 1

The correct option is B $\frac{x^2}{a^2}+\frac{y^2}{b^2}=2$

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

Adding (i) and (ii), we get,

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