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  • The area of the triangle inscribed in an ellipse is <img alt="2ab \sin \frac{\beta-\gamma}{2} \cdot \sin \frac{\gamma-\alpha}{2} \cdot \sin \frac{\alpha-\beta}{2}, \text{where}\: \alpha,

The area of the triangle inscribed in an ellipse is 2ab \sin \frac{\beta-\gamma}{2} \cdot \sin \frac{\gamma-\alpha}{2} \cdot \sin \frac{\alpha-\beta}{2}, \text{where}\: \alpha, \beta\; \text{and} \: \gamma are the eccentric angles of the vertices, then the condition

Option: 1

\alpha, \alpha-\frac{ \pi}{3} \: \text { and } \alpha+\frac{4 \pi}{3}


Option: 2

2\alpha, \alpha+\frac{2 \pi}{3} \: \text { and } \alpha+\frac{4 \pi}{3}


Option: 3

\alpha, \alpha+\frac{2 \pi}{3} \: \text { and } \alpha+\frac{4 \pi}{3}

 


Option: 4

\alpha, \alpha-\frac{2 \pi}{3} \: \text { and } \alpha+\frac{4 \pi}{3}


Answers (1)

best_answer

(b) Let P, Q and R be the angular points of the triangle and let \alpha, \beta\: \text{and}\: \gamma be their eccentric angles, respectively.
The vertices of the \triangle P Q R\: \text{and}\: (a \cos \alpha, b \sin \alpha), (a \cos \beta, b \sin \beta) \: \text{and}\: (a \cos \gamma, b \cos \gamma).
Therefore, the area of the \triangle P Q R

                  =2 a b \sin \frac{\beta-\gamma}{2} \sin \frac{\gamma-\alpha}{2} \sin \frac{\alpha-\beta}{2}

Let p, q and r be the corresponding points on the auxiliary circle.

Then, area of \Delta p q r=2 a^2 \sin \frac{\beta-\gamma}{2} \sin \frac{\gamma-\alpha}{2} \sin \frac{\alpha-\beta}{2}

On putting b = a , we get

                   \begin{aligned} & \frac{\operatorname{ar}(\triangle P Q R)}{\operatorname{ar}(\triangle p q r)}=\frac{b}{a} \\ \\&\Rightarrow\: \: \: \: \: \: \: \: \: \: \: \: \operatorname{ar}(\triangle P Q R)=\frac{b}{a} \operatorname{ar}(\triangle p q r) \end{aligned}

Hence, \triangle P Q R will be maximum when \triangle p q r is maximum but \Delta p q r is maximum when it is an equilateral triangle and in that case,

                  \alpha-\beta=\beta-\gamma=\gamma-\alpha=\frac{2 \pi}{3}

Hence, when a triangle inscribed in an ellipse is maximum, the eccentric angles of its angular points are \alpha, \alpha+\frac{2 \pi}{3} \: \text { and } \alpha+\frac{4 \pi}{3}

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