Question

Prove that the area of the triangle formed by lines joining the points on the ellipse

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

whose eccentric angles are

α, β, γ

is

\frac{1}{2} a b \{\sin (β - γ) + \sin (α - β) + \sin (γ - α)\} .

Accepted Answer

We know that, the co-ordinates of a point on the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ with eccentric angle $θ$ is $(a \cos θ, b \sin θ) .$

Thus, the points on the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ whose eccentric angles are $α, β, γ$ are $(a \cos α, b \sin α), (a \cos β, b \sin β)$ and $(a \cos γ, b \sin γ) .$

The area of the triangle with vertices $(x_{1}, y_{1}), (x_{2}, y_{2})$ and $(x_{3}, y_{3})$ is the absolute value of $\frac{1}{2} |\begin{matrix} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{matrix}| .$

Thus, the area of the required triangle is $∆ = \frac{1}{2} |\begin{matrix} a \cos α & b \sin α & 1 \\ a \cos β & b \sin β & 1 \\ a \cos γ & b \sin γ & 1 \end{matrix}| .$

Taking $a$ common from the first column and $b$ from the second column.

$\Rightarrow ∆ = \frac{a b}{2} |\begin{matrix} \cos α & \sin α & 1 \\ \cos β & \sin β & 1 \\ \cos γ & \sin γ & 1 \end{matrix}| .$

Applying $R_{2} \to R_{2} - R_{1}, R_{3} \to R_{3} - R_{1}$

$\Rightarrow ∆ = \frac{1}{2} a b |\begin{matrix} \cos α & \sin α & 1 \\ \cos β - \cos α & \sin β - \sin α & 0 \\ \cos γ - \cos α & \sin γ - \sin α & 0 \end{matrix}|$

$\Rightarrow ∆ = \frac{1}{2} a b |(\cos β - \cos α) (\sin γ - \sin α) - (\sin β - \sin α) (\cos γ - \cos α)|$

$\Rightarrow ∆ = \frac{1}{2} a b |\cos β \sin γ - \cos α \sin γ - \cos β \sin α + \cos α \sin α - \sin β \cos γ + \sin α \cos γ + \sin β \cos α - \sin α \cos α|$

$\Rightarrow ∆ = \frac{1}{2} a b |\cos β \sin γ - \cos α \sin γ - \cos β \sin α - \sin β \cos γ + \sin α \cos γ + \sin β \cos α|$

$\Rightarrow ∆ = \frac{1}{2} a b |(\cos β \sin γ - \sin β \cos γ) + (\sin α \cos γ - \cos α \sin γ) + (\sin β \cos α - \cos β \sin α)| .$

Using $\sin A \cos B - \cos A \sin B = \sin (A - B),$ we get

$∆ = \frac{1}{2} a b |\sin (γ - β) + \sin (α - γ) + \sin (β - α)| .$

Also, we know that $\sin (- θ) = - \sin θ .$

$\Rightarrow ∆ = \frac{1}{2} a b |\sin (β - γ) + \sin (α - β) + \sin (γ - α)| .$

Dr. SK Goyal Solutions for Chapter: Ellipse, Exercise 5: EXERCISE ON LEVEL-II

Dr. SK Goyal Mathematics Solutions for Exercise - Dr. SK Goyal Solutions for Chapter: Ellipse, Exercise 5: EXERCISE ON LEVEL-II

Questions from Dr. SK Goyal Solutions for Chapter: Ellipse, Exercise 5: EXERCISE ON LEVEL-II with Hints & Solutions

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