Question

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

Answer 1

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

Squaring both side, we have:

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

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

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

𝑆𝑞𝑢𝑎𝑟𝑖𝑛𝑔 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒, 𝑤𝑒 𝑔𝑒𝑡:

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

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

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

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

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

 ⇒${\displaystyle (\frac{x^2}{a^2}{\sin }^2\theta +\frac{x^2}{a^2}{\cos }^2\theta )+(\frac{y^2}{b^2}{\cos }^2\theta +\frac{y^2}{b^2}{\sin }^2\theta )=2}$

 ⇒${\displaystyle \frac{x^2}{a^2}({\sin }^2\theta +{\cos }^2\theta )+\frac{y^2}{b^2}({\cos }^2\theta +{\sin }^2\theta )=2}$

 ⇒${\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=2 [\because {\sin }^2\theta +{\cos }^2\theta =1]}$

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