Question : If $x\cos \theta -y\sin \theta =\sqrt{x^{2}+y^{2}}$ and $\frac{\cos ^2{\theta }}{a^{2}}+\frac{\sin ^{2}\theta}{b^{2}}=\frac{1}{x^{2}+y^{2}},$ then the correct relation is:
Option 1: $\frac{x^{2}}{b^{2}}-\frac{y^{2}}{a^{2}}=1$
Option 2: $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$
Option 3: $\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1$
Option 4: $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$
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Correct Answer: $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$
Solution :
$x\cos \theta -y\sin \theta =\sqrt{x^{2}+y^{2}}$ ..................................(1)
$\frac{\cos ^2{\theta }}{a^{2}}+\frac{\sin ^{2}\theta }{b^{2}}=\frac{1}{x^{2}+y^{2}}$ .......................................(2)
From equation (1),
$\frac{x\cos \theta}{\sqrt{x^{2}+y^{2}}} -\frac{y\sin \theta}{\sqrt{x^{2}+y^{2}}} =1$
Comparing with $\cos^2 \theta + \sin^2 \theta = 1$, we get,
$\sin \theta = -\frac{y}{\sqrt{x^{2}+y^{2}}}$ and $\cos \theta = \frac{x}{\sqrt{x^{2}+y^{2}}}$
Putting in (2),
$\frac{x^2}{(x^2 + y^2)a^2}+ \frac{y^2}{(x^2 + y^2)b^2} = \frac{1}{(x^2 + y^2)}$
⇒ $\frac{x^2}{a^2}+ \frac{y^2}{b^2} = 1$
Hence, the correct answer is $\frac{x^2}{a^2}+ \frac{y^2}{b^2} = 1$.
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