If \(2 y \cos \theta=x \sin \theta\) and \(2 x \sec \theta-y \operatorname{cosec} \theta=3\), then \(x^2+4…
If \(2 y \cos \theta=x \sin \theta\) and \(2 x \sec \theta-y \operatorname{cosec} \theta=3\), then \(x^2+4 y^2=\)
2
4
1
none
Solution
We have \(2 \mathrm{y} \cos \theta=\mathrm{x} \sin \theta\)
or \(\frac{\cos \theta}{\mathrm{x}}=\frac{\sin \theta}{2 \mathrm{y}}=\mathrm{k}\) (say)
Then \(\cos \theta=\mathrm{kx}\) and \(\sin \theta=2 \mathrm{ky}\) ...(1)
Again \(2 \mathrm{x} \sec \theta-\mathrm{y} \operatorname{cosec} \theta=3\)
or \(\frac{2 \mathrm{x}}{\cos \theta}-\frac{\mathrm{y}}{\sin \theta}=3\)
or \(\frac{2 \mathrm{x}}{\mathrm{kx}}-\frac{\mathrm{y}}{2 \mathrm{ky}}=3 \quad\) [from(1)]
or \(\frac{2}{\mathrm{k}}-\frac{1}{2 \mathrm{k}}=3\), giving \(\mathrm{k}=\frac{1}{2}\)
We now get \(\cos \theta=\frac{\mathrm{x}}{2}\) and \(\sin \theta=\mathrm{y}\)
Squaring and adding we get
\(\begin{aligned}
& \cos ^2 \theta+\sin ^2 \theta=\frac{x^2}{4+y^2} \\
& \Rightarrow \frac{x^2}{4}+y^2=1, \text { or } x^2+4 y^2=4
\end{aligned}\)