AP EAMCET · Maths · Ellipse
If \((l, m)\) is the circumcentre of an equilateral triangle inscribed in the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) having vertices at points with eccentric angles \(\theta_1, \theta_2\) and \(\theta_3\), then \(\frac{2}{3}\left[\cos \left(\theta_1-\theta_2\right)+\cos \left(\theta_2-\theta_3\right)+\cos \left(\theta_3-\theta_1\right)\right]=\)
- A \(\frac{9 l^2}{2 a^2}+\frac{9 m^2}{b^2}-1\)
- B \(\frac{l^2}{\mathrm{a}^2}+\frac{m^2}{b^2}-3\)
- C \(\frac{3 l^2}{a^2}+\frac{3 m^2}{b^2}-1\)
- D \(\frac{3 l^2}{a^2}+\frac{3 m^2}{b^2}-\frac{3}{2}\)
Answer & Solution
Correct Answer
(C) \(\frac{3 l^2}{a^2}+\frac{3 m^2}{b^2}-1\)
Step-by-step Solution
Detailed explanation
As we know coordinates of circumcentre of an equilateral triangle is same as centroid, so…
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