Question

If $\frac{x}{a}cos\theta +\frac{y}{b}sin\theta =1$ and $\frac{x}{a}sin\theta -\frac{y}{b}cos\theta =1,$ then which of the following equations is true ?

• A. $\frac{x^2}{a^2}+\frac{y^2}{b^2}=0$
• B. $\frac{x^2}{a^2}+\frac{y^2}{b^2}=2$
• C. $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
• D. $\frac{x^2}{a^2}-\frac{y^2}{b^2}=2$

Answer 1

The correct option is B

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

Given that $\frac{x}{a}cos\theta +\frac{y}{b}sin\theta =1,\frac{x}{a}sin\theta -\frac{y}{b}cos\theta =1$.

Squaring the given equations we get,
$(\frac{x}{a}cos\theta +\frac{y}{b}sin\theta {)}^2=1,(\frac{x}{a}sin\theta -\frac{y}{b}cos\theta {)}^2=1$

Then, $\frac{x^2}{a^2}co{s}^2\theta +\frac{2xy}{ab}cos\theta sin\theta +\frac{y^2}{b^2}si{n}^2\theta =1$-----(i)
and $\frac{x^2}{a^2}si{n}^2\theta -\frac{2xy}{ab}sin\theta cos\theta +\frac{y^2}{b^2}co{s}^2\theta =1$-----(ii)

Adding these two equations (i) and (ii), we get
$\frac{x^2}{a^2}si{n}^2\theta -\frac{2xy}{ab}cos\theta sin\theta +\frac{y^2}{b^2}co{s}^2\theta +\frac{x^2}{a^2}co{s}^2\theta +\frac{2xy}{ab}cos\theta sin\theta +\frac{y^2}{b^2}si{n}^2\theta =2$

On simplifying this equation, we get
$\frac{x^2}{a^2}(si{n}^2\theta +co{s}^2\theta )+\frac{y^2}{b^2}(co{s}^2\theta +si{n}^2\theta )=2$

$\therefore \frac{x^2}{a^2}+\frac{y^2}{b^2}=2(\because si{n}^2\theta +co{s}^2\theta =1)$