WBJEE · Maths · Ellipse
The eccentric angle in the first quadrant of a point on the ellipse \(\frac{x^{2}}{10}+\frac{y^{2}}{8}=1\) at a distance 3 units from the centre of the ellipse is
- A \(\frac{\pi}{6}\)
- B \(\frac{\pi}{4}\)
- C \(\frac{\pi}{3}\)
- D \(\frac{\pi}{2}\)
Answer & Solution
Correct Answer
(B) \(\frac{\pi}{4}\)
Step-by-step Solution
Detailed explanation
Let \(P(\sqrt{10} \cos \theta, \sqrt{8} \sin \theta)\) be the required point on \(\frac{x^{2}}{10}+\frac{y^{2}}{8}=1\) Whose distance from centre (0,0) is 3 units. \(\therefore 10 \cos ^{2} \theta+8 \sin ^{2} \theta=9=9\left(\sin ^{2} \theta+\cos ^{2} \theta\right)\)…
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