Conic Sections Question 317
Question: A point on the ellipse $ \frac{x^{2}}{16}+\frac{y^{2}}{9}=1 $ at a distance equal to the mean of the lengths of the semi-major axis and semi-minor axis from the centre is
Options:
A) $ ( \frac{2\sqrt{91}}{7},\frac{3\sqrt{105}}{14} ) $
B) $ ( \frac{2\sqrt{91}}{7}-\frac{3\sqrt{105}}{14} ) $
C) $ ( \frac{2\sqrt{105}}{7},\frac{3\sqrt{91}}{14} ) $
D) $ ( -\frac{2\sqrt{105}}{7}-\frac{3\sqrt{91}}{14} ) $
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Answer:
Correct Answer: A
Solution:
[a] Let the point is $ (4cos\theta ,3sin\theta ) $ According to question, $ {{(4cos)}^{2}}+{{(3sin\theta )}^{2}}={{( \frac{4+3}{2} )}^{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 $ ( \frac{2\sqrt{91}}{7},\frac{3\sqrt{105}}{14} ); $
$ ( -\frac{2\sqrt{91}}{7},-\frac{3\sqrt{105}}{14} ) $ Interchange $ \theta $ by $ \frac{\pi }{2}+\theta $ and $ \frac{3\pi }{2}+\theta $
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