Question

Given $2y\cos \theta =x\sin \theta and2x\sec \theta -ycosec\theta =3$ then the value of $x^2+4y^2$ is equal to

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is D 4

Given: $2y\cos \theta =x\sin \theta$ (1)
$2x\sec \theta -ycosec\theta =3$
$2x\sin \theta -y\cos \theta =3\sin \theta \cos \theta$
$2\left(2y\cos \theta \right)-y\cos \theta =3\sin \theta \cos \theta$
$3y=3\sin \theta$
$\sin \theta =y$
Put this value in equation 1
Thus, $2y\cos \theta =x\sin \theta$
$2\sin \theta \cos \theta =x\sin \theta$
$\cos \theta =\frac{x}{2}$
We know, ${\sin }^2\theta +{\cos }^2\theta =1$
$y^2+(\frac{x}{2}{)}^2=1$
$4y^2+x^2=4$