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 eliminate $\theta$

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

Answer 1

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

Given, $\frac{x}{a}\cos \theta +\frac{y}{b}\sin \theta =1,\frac{x}{a}\sin \theta -\frac{y}{b}\cos \theta =1$
Squaring and adding them
$\Rightarrow \frac{x^2}{a^2}{\cos }^2\theta +\frac{y^2}{b^2}{\sin }^2\theta +\frac{2xy}{ab}\sin \theta \cos \theta +$
$\frac{x^2}{a^2}{\sin }^2\theta +\frac{y^2}{b^2}{\cos }^2\theta -\frac{2xy}{ab}\sin \theta \cos \theta =1+1$
$\Rightarrow \frac{x^2}{a^2}+\frac{y^2}{b^2}=2$