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If orthocentre of an equilateral triangle, inscribed in the ellipse \mathrm{\frac{x^2}{a^2}+\frac{y^2}{b^2}=1}with vertices having eccentric angles \mathrm{\alpha, \beta, \gamma} respectively, is \mathrm{(l, m)} then \mathrm{\Sigma \cos (\alpha-\beta)} equals

Option: 1

\mathrm{9 l^2+9 m^2+2 a^2 b^2}


Option: 2

\mathrm{\frac{9 l^2}{2 a^2}+\frac{9 m^2}{2 b^2}-\frac{3}{2}}


Option: 3

\mathrm{\frac{9 l^2}{2 a^2}+\frac{9 m^2}{2 b^2}-\frac{3}{2} a^2 b^2}


Option: 4

\mathrm{\frac{9 l^2}{a^2}+\frac{9 m^2}{b^2}+\frac{3}{2}}


Answers (1)

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We know that in an equilateral triangle, incentre, orthocentre, circumcentre and centroid coincide with each other. Therefore, orthocentre of triangle having vertices \mathrm{A(a \cos \alpha, b \sin \alpha), B(a \cos \beta, b \sin \beta), C(a \cos \gamma, b \sin \gamma)} be

\mathrm{ \begin{aligned} & (l, m)=\left(\frac{a(\cos \alpha+\cos \beta+\cos \gamma)}{3}, \frac{b(\sin \alpha+\sin \beta+\sin \gamma)}{3}\right) \\ & \Rightarrow \quad(\cos \alpha+\cos \beta+\cos \gamma)=\frac{3 l}{a}\, \, \, \, ...(1) \\ & \text { and }(\sin \alpha+\sin \beta+\sin \gamma)=\frac{3 m}{b} \, \, \, .....(2)\end{aligned} }
Now, squaring (1) and (2) and then adding, we get \mathrm{(\cos \alpha+\cos \beta+\cos \gamma)^2+(\sin \alpha+\sin \beta+\sin \gamma)^2 =\frac{9 l^2}{a^2}+\frac{9 m^2}{b^2}}

\mathrm{\Rightarrow \quad\left(\cos ^2 \alpha+\sin ^2 \alpha\right)+\left(\cos ^2 \beta+\sin ^2 \beta\right)+\left(\cos ^2 \gamma+\sin ^2 \gamma\right)}

                                                                   \mathrm{+2 \Sigma \cos (\alpha-\beta)=\frac{9 l^2}{a^2}+\frac{9 m^2}{b^2}}

\mathrm{\Rightarrow \quad 2 \Sigma \cos (\alpha-\beta)=\frac{9 l^2}{a^2}+\frac{9 m^2}{b^2}-3}

\mathrm{\Rightarrow \quad \Sigma \cos (\alpha-\beta)=\frac{9 l^2}{2 a^2}+\frac{9 m^2}{2 b^2}-\frac{3}{2}}

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Nehul

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