# Solve this

Question:

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

Solution:

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

Squaring both side, we have:

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

$\Rightarrow\left(\frac{x^{2}}{a^{2}} \sin ^{2} \theta+\frac{y^{2}}{b^{2}} \cos ^{2} \theta-2 \frac{x}{a} \times \frac{y}{b} \sin \theta \cos \theta\right)=1$    ...........(i)

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

Squaring both side, we get:

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

$\Rightarrow\left(\frac{x^{2}}{a^{2}} \cos ^{2} \theta+\frac{y^{2}}{b^{2}} \sin ^{2} \theta+2 \frac{x}{a} \times \frac{y}{b} \sin \theta \cos \theta\right)=1$   ............(ii)

Now, adding (i) and (ii), we get:

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

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

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

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

$\Rightarrow \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=2 \quad\left[\because \sin ^{2} \theta+\cos ^{2} \theta=1\right]$

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