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  • The area of a triangle inscribed in an ellipse bears a constant ratio to the area of the triangle formed by joining points on the auxiliary circle corresponding to the vertices of the fir

The area of a triangle inscribed in an ellipse bears a constant ratio to the area of the triangle formed by joining points on the auxiliary circle corresponding to the vertices of the first triangle. This ratio is

Option: 1

\frac{b}{a}


Option: 2

\frac{a^2}{b^2}


Option: 3

\frac{2a}{b}

 


Option: 4

None


Answers (1)

best_answer

(a) Let P\left(a \cos \theta_1, b \sin \theta_1\right), \quad Q\left(a \cos \theta_2, \quad b \sin \theta_2\right) and R\left(a \cos \theta_3, b \sin \theta_3\right) be the vertices of the triangle

inscribe in the ellipse \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 . The points on the auxiliary circle corresponding to these points are \begin{aligned} & P^{\prime}\left(a \cos \theta_1, a \sin \theta_1\right), Q^{\prime}\left(a \cos \theta_2, a \sin \theta_2\right) \text { and } \\ & R^{\prime}\left(a \cos \theta_3, a \sin \theta_3\right) \end{aligned} 

\quad \begin{aligned} \therefore \: \: \: \: \: \: \: \: \: \Delta_1 & =\text { Area of } \triangle P Q R \\ & =\frac{1}{2}\left|\begin{array}{lll} a \cos \theta_1 & b \sin \theta_1 & 1 \\ a \cos \theta_2 & b \sin \theta_2 & 1 \\ a \cos \theta_3 & b \sin \theta_3 & 1 \end{array}\right| \\ & =\frac{1}{2} a b\left|\begin{array}{lll} \cos \theta_1 & \sin \theta_1 & 1 \\ \cos \theta_2 & \sin \theta_2 & 1 \\ \cos \theta_3 & \sin \theta_3 & 1 \end{array}\right| \end{aligned}

\text { and } \Delta_2=\text { Area of } \Delta P^{\prime} Q^{\prime} R^{\prime}

                     \begin{aligned} & =\frac{1}{2}\left|\begin{array}{lll} a \cos \theta_1 & a \sin \theta_1 & 1 \\ a \cos \theta_2 & a \sin \theta_2 & 1 \\ a \cos \theta_3 & a \sin \theta_3 & 1 \end{array}\right| \\ \\& =\frac{1}{2} a^2\left|\begin{array}{lll} \cos \theta_1 & \sin \theta_1 & 1 \\ \cos \theta_2 & \sin \theta_2 & 1 \\ \cos \theta_3 & \sin \theta_3 & 1 \end{array}\right| \end{aligned}

\text { Clearly, } \frac{\Delta_1}{\Delta_2}=\frac{b}{a}=\text { constant }

Posted by

SANGALDEEP SINGH

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