Question

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-

• A. b/a
• B. a²/b²
• C. 2a/b
• D. None of these

Answer 1

Answer: (A)

b/a

Answer 2

Let P(a cos q₁, b sin q₁), Q(a cos q₂, b sin q₂) and R(a cos q₃, b sin q₃) be the vertices of the triangle inscribed in the ellipse x²/a² + y²/b² = 1. The points on the auxiliary circle corresponding to these points are P¢(a cos q₁, a sin q₁), Q¢(a cos q₂ , a sin q₂) and R¢(a cos q₃, a sin q₃).

\ D₁ = Area of DPQR

= $\frac{1}{2}\left| \begin{matrix} a\cos\theta_{1} & b\sin\theta_{1} & 1 \\ a\cos\theta_{2} & b\sin\theta_{2} & 1 \\ a\cos\theta_{3} & b\sin\theta_{3} & 1 \end{matrix} \right|$

= $\frac{ab}{2}\left| \begin{matrix} \cos\theta_{1} & \sin\theta_{1} & 1 \\ \cos\theta_{2} & \sin\theta_{2} & 1 \\ \cos\theta_{3} & \sin\theta_{3} & 1 \end{matrix} \right|$

and, D₂ = Area of DP¢ Q¢ R¢

=$\frac{1}{2}\left| \begin{matrix} a\cos\theta_{1} & b\sin\theta_{1} & 1 \\ a\cos\theta_{2} & b\sin\theta_{2} & 1 \\ a\cos\theta_{3} & b\sin\theta_{3} & 1 \end{matrix} \right|$

= $\frac{1}{2}a^{2}\left| \begin{matrix} \cos\theta_{1} & \sin\theta_{1} & 1 \\ \cos\theta_{2} & \sin\theta_{2} & 1 \\ \cos\theta_{3} & \sin\theta_{3} & 1 \end{matrix} \right|$

Clearly, $\frac{\Delta_{1}}{\Delta_{2}}$ = $\frac{b}{a}$ = constant.