A) \[\left( \frac{2\sqrt{91}}{7},\frac{3\sqrt{105}}{14} \right)\]
B) \[\left( \frac{2\sqrt{91}}{7}-\frac{3\sqrt{105}}{14} \right)\]
C) \[\left( \frac{2\sqrt{105}}{7},\frac{3\sqrt{91}}{14} \right)\]
D) \[\left( -\frac{2\sqrt{105}}{7}-\frac{3\sqrt{91}}{14} \right)\]
Correct Answer: A
Solution :
[a] Let the point is \[(4cos\theta ,3sin\theta )\] According to question, \[{{(4cos)}^{2}}+{{(3sin\theta )}^{2}}={{\left( \frac{4+3}{2} \right)}^{2}}\] ? (1) From (1) \[16-7{{\sin }^{2}}\theta =\frac{49}{4}\Rightarrow {{\sin }^{2}}\theta =\frac{15}{28}\] \[\therefore \sin \theta =\pm \frac{1}{2}\sqrt{\frac{15}{7}}=\pm \frac{\sqrt{105}}{14}\] Similarly, \[\cos \theta =\pm \frac{\sqrt{91}}{14}\] So the points are \[\left( \frac{2\sqrt{91}}{7},\frac{3\sqrt{105}}{14} \right);\] \[\left( -\frac{2\sqrt{91}}{7},-\frac{3\sqrt{105}}{14} \right)\] Interchange \[\theta \] by \[\frac{\pi }{2}+\theta \] and \[\frac{3\pi }{2}+\theta \]
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