# This is the question ?

## $\frac{x}{a} \cos \theta + \frac{y}{b} \sin \theta = 1$ $\frac{x}{a} \sin \theta - \frac{y}{b} \cos \theta = 1$ Then prove that ${x}^{2} / {a}^{2} + {y}^{2} / {b}^{2} = 2$

Jun 1, 2018

Given,

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

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

Squaring and adding equation $\left(1\right)$ and $\left(2\right)$,we get

$\rightarrow \frac{{x}^{2}}{{a}^{2}} {\cos}^{2} \theta \cancel{+ 2 \left(\frac{x y}{a b}\right) \cos \theta \cdot \sin \theta} + \frac{{y}^{2}}{{b}^{2}} {\sin}^{2} \theta + \frac{{x}^{2}}{{a}^{2}} {\sin}^{2} \theta \cancel{- 2 \left(\frac{x y}{a b}\right) \sin \theta \cdot \cos \theta} + \frac{{y}^{2}}{{b}^{2}} {\cos}^{2} \theta = {1}^{2} + {1}^{2}$

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

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

Jun 1, 2018

#### Explanation:

Here,

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

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

Squaring $\left(1\right) \mathmr{and} \left(2\right) \mathmr{and} t h e n$ adding we get

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

ul(x^2/a^2sin^2theta- 2(xy)/(ab)costhetasintheta+y^2/b^2cos^2theta=1)

${x}^{2} / {a}^{2} \left({\cos}^{2} \theta + {\sin}^{2} \theta\right) + 0 + {y}^{2} / {b}^{2} \left({\sin}^{2} \theta + {\cos}^{2} \theta\right) = 1 + 1$

$i . e . {x}^{2} / {a}^{2} \left(1\right) + {y}^{2} / {b}^{2} \left(1\right) = 2$

Hence,

${x}^{2} / {a}^{2} + {y}^{2} / {b}^{2} = 2$