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If circumcentre of an equilateral triangle inscribed in $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ , with vertices having eccentric angles $\alpha ,\beta ,\gamma$ respectively is (x₁, y₁), then $\sum cos\;\alpha \; cos \;\beta + \sum sin \; \alpha \; sin \;\beta =$
Option: 1
$\frac{9x_{1}^{2}}{a^{2}}+\frac{9y_{1}^{2}}{b^{2}}+\frac{3}{2}$

Option: 2
$9x_{1}^{2}-9y_{1}^{2}+a^{2}b^{2}$

Option: 3
$\frac{9x_{1}^{2}}{a}+\frac{9y_{1}^{2}}{b}+3$

Option: 4
$\frac{9x_{1}^{2}}{2a^{2}}+\frac{9y_{1}^{2}}{2b^{2}}-\frac{3}{2}$

Answers (1)

best_answer

Standard equation -

$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}= 1$

- wherein

$a\rightarrow$ Semi major axis

$b\rightarrow$ Semi minor axis

$x_{1}=\frac{a}{3}(cos \alpha +cos \beta +cos \gamma )$

$y_{1}=\frac{b}{3}(sin \alpha +sin \beta +sin \gamma )$

$\frac{x_{1}^{2}}{a^{2}}+\frac{y_{1}^{2}}{b^{2}}$

$=\left ( \frac{cos\alpha +cos\beta +cos\gamma }{3} \right )^{2}+\left ( \frac{sin\alpha +sin\beta +sin\gamma }{3} \right )^{2}$

$\Rightarrow \frac{9x_{1}^{2}}{a^{2}}+ \frac{9y_{1}^{2}}{b^{2}}$

$=1+1+1+2\left ( cos\alpha \; cos\beta +cos\beta \; cos\gamma+cos\gamma \;cos\alpha + sin\alpha \; sin\beta +sin\beta \; sin\gamma+sin\gamma \;sin\alpha \right )$

$\Rightarrow \sum cos\alpha \:cos\beta + \sum sin\alpha \:sin\beta=\frac{9x_{1}^{2}}{2a^{2}}+\frac{9y_{1}^{2}}{2b^{2}}-\frac{3}{2}$

Posted by

Ritika Harsh

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