A) 1
B) 2
C) 3
D) 4
Correct Answer: D
Solution :
| [d] Given, \[2y\cos \theta =x\sin \theta \] ...(i) \[\Rightarrow \] \[2y\cdot \frac{1}{\sec \theta }=\frac{x}{\text{cosec}\theta }\] \[\Rightarrow \] \[x\sec \theta -2y\cos ec\theta =0\] and \[2x\sec \theta -y\,\text{cosec}\theta =3\] ...(iii) On solving Eqs. (i) and (ii), we get \[y=\sin \theta \] From Eq. (i), we get \[x=2\cos \theta \] \[\therefore \] \[{{x}^{2}}+4{{y}^{2}}={{(2cos\theta )}^{2}}+4\,{{(sin\theta )}^{2}}\] \[=4\,(co{{s}^{2}}\theta +si{{n}^{2}}\theta )=4\] |
You need to login to perform this action.
You will be redirected in
3 sec
