Question

If $\frac{x}{a}$ cosФ + $\frac{y}{b}$ sin Ф = 1 and $\frac{x}{a}$ sin Ф - $\frac{y}{b}$ cos Ф = 1
Prove that $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 2$

Answer 1

From first equation;
$\frac{x}{a}\ cos \phi+ \frac{y}{b}\ sin \phi=1\\ \\ \Rightarrow (\frac{x}{a}\ cos \phi+ \frac{y}{b}\ sin \phi)^2=1^2\\ \\ \Rightarrow \frac{x^2}{a^2}\ cos^2 \phi+ \frac{y^2}{b^2}\ sin^2 \phi+2 \times \frac{x}{a}\ cos \phi \times \frac{y}{b}\ sin \phi=1\\ \\ \Rightarrow\frac{x^2}{a^2}\ cos^2 \phi+ \frac{y^2}{b^2}\ sin^2 \phi+ \frac{2xy}{ab}\ cos \phi\ sin \phi=1$

From second equation;
$\frac{x}{a}\ sin \phi- \frac{y}{b}\ cos \phi=1\\ \\ \Rightarrow (\frac{x}{a}\ sin\phi- \frac{y}{b}\ cos \phi)^2=1^2\\ \\ \Rightarrow \frac{x^2}{a^2}\ sin^2 \phi+ \frac{y^2}{b^2}\ cos^2 \phi-2 \times \frac{x}{a}\ sin \phi \times \frac{y}{b}\ cos \phi=1\\ \\ \Rightarrow\frac{x^2}{a^2}\ sin^2 \phi+ \frac{y^2}{b^2}\ cos^2 \phi- \frac{2xy}{ab}\ sin \phi\ cos \phi=1$

Adding both the equations, we get 
$\frac{x^2}{a^2}\ cos^2 \phi+ \frac{y^2}{b^2}\ sin^2 \phi+\frac{x^2}{a^2}\ sin^2 \phi+ \frac{y^2}{b^2}\ cos^2 \phi=1+1\\ \\ \Rightarrow \frac{x^2}{a^2}\ cos^2 \phi+\frac{x^2}{a^2}\ sin^2 \phi+ \frac{y^2}{b^2}\ sin^2 \phi+ \frac{y^2}{b^2}\ cos^2 \phi=2\\ \\ \Rightarrow \frac{x^2}{a^2}\ (cos^2 \phi + sin^2 \phi)+ \frac{y^2}{b^2}\ (sin^2 \phi+cos^2 \phi)=2\\ \\ \Rightarrow \frac{x^2}{a^2}\ (1)+ \frac{y^2}{b^2}\ (1)=2\\ \\ \Rightarrow \frac{x^2}{a^2}\ + \frac{y^2}{b^2}\ =2$