Question

If $2y\cos \theta =x\sin \theta$ and $2xsec\theta -y\cos ec\theta =3$, then $x^2+4y^2=$

• A. $4$
• B. $-4$
• C. $\pm 4$
• D. None of these

Answer 1

The correct option is A

$4$

Explanation for the correct option:

Step 1: Find the value of $x^2+4y^2$:

Given,

$2y\cos \theta =x\sin \theta$ ….(i)

Squaring on both sides,

$4y^2{\cos }^2\theta =x^2{\sin }^2\theta$ ….(ii)

Step 2. Again from the given, we get

$2xsec\theta –y\cos ec\theta =3$

$\Rightarrow$ $\frac{2x}{\cos \theta }–\frac{y}{\sin \theta }=3$

$\Rightarrow$ $2x\sin \theta –y\cos \theta =3\sin \theta \cos \theta$

$\Rightarrow$$2x\sin \theta –\frac{x\sin \theta }{2}=3\sin \theta \cos \theta$ $\left[\because Fromi\right]$

$\Rightarrow$ $3x\sin \theta =6\sin \theta \cos \theta$

$\Rightarrow$ $x=2\cos \theta$ ….(iii)

Step 3. From (ii) and (iii), we get

$\begin{array}{rcl}{x}^2+4y^2& =& {(2\cos \theta )}^2+\left[\frac{(x^2{\sin }^2\theta )}{{\cos }^2\theta }\right]\\ & =& 4{\cos }^2\theta +\frac{(4{\cos }^2\theta ){\sin }^2\theta }{{\cos }^2\theta }\\ & =& 4({\cos }^2\theta +{\sin }^2\theta )\mathbf{}\left[\because co{s}^2\theta +si{n}^2\theta =1\right]\\ & =& \mathbf{4}\end{array}$

Hence, Option $A$ is Correct.