TS EAMCET · Maths · Ellipse
The circumcenter of the equilateral triangle having the three points \(\theta_1, \theta_2, \theta_3\) lying on the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) as its vertices is \((r, s)\). Then the average of \(\cos \left(\theta_1-\theta_2\right)\), \(\cos \left(\theta_2-\theta_3\right)\) and \(\cos \left(\theta_3-\theta_1\right)\) is
- A \(\frac{1}{2}\left[\frac{3 r^2}{a^2}+\frac{3 s^2}{b^2}-1\right]\)
- B \(\frac{3}{2}\left[\frac{r^2}{a^2}+\frac{s^2}{b^2}\right]\)
- C \(\frac{1}{3}\left[\frac{r^2}{a^2}+\frac{s^2}{b^2}\right]\)
- D \(\frac{1}{3}\left[\frac{r^2}{a^2}+\frac{s^2}{b^2}+\frac{r s}{a b}\right]\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{2}\left[\frac{3 r^2}{a^2}+\frac{3 s^2}{b^2}-1\right]\)
Step-by-step Solution
Detailed explanation
\(r = \frac{a}{3}\sum \cos \theta_i\) \(s = \frac{b}{3}\sum \sin \theta_i\) \(\frac{r^2}{a^2} + \frac{s^2}{b^2} = \frac{1}{9} \left[ (\sum \cos \theta_i)^2 + (\sum \sin \theta_i)^2 \right]\)…
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