Three points A - B - C are taken on an ellipse with eccentric angles - - and -2 - respectively- The area of - ABC isA- independent of -B- minimum for -2 -3C- maximum for -214-3D- dependent on -

The correct options are A independent of θ C maximum for α=2π3Let Δ be the area of the triangle whose vertices are A(acosθ,bsinθ), B(acos (θ+α),bsin (θ+α)) and ...

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Standard XII

Mathematics

Area of Triangle Using Coordinates

Three points ...

Question

Three points $A, B, C$ are taken on an ellipse with eccentric angles $θ, θ + α$ and $θ + 2 α$ respectively. The area of $Δ A B C$ is

independent of

θ

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minimum for

α = \frac{2 π}{3}

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maximum for

α = \frac{2 π}{3}

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dependent on

θ

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Solution

The correct options are
A independent of $θ$
C maximum for $α = \frac{2 π}{3}$
Let $Δ$ be the area of the triangle whose vertices are $A (a cos θ, b sin θ), B (a cos (θ + α), b sin (θ + α))$
and $C (a cos (θ + 2 α), b sin (θ + 2 α))$ .

We have,
$Δ = \frac{1}{2} ∣ ∣ ∣ ∣ ∣ \begin{matrix} a cos θ & b sin θ & 1 a cos (θ + α) & b sin (θ + α) & 1 a cos (θ + 2 α) & b sin (θ + 2 α) & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣$

$R_{2} \to R_{2} - R_{1}$ and $R_{3} \to R_{3} - R_{1}$
$= \frac{a b}{2} ∣ ∣ ∣ ∣ ∣ \begin{matrix} cos θ & sin θ & 1 cos (θ + α) - cos θ & sin (θ + α) - sin θ & 0 cos (θ + 2 α) - cos θ & sin (θ + 2 α) - sin θ & 0 \end{matrix} ∣ ∣ ∣ ∣ ∣$

$= \frac{a b}{2} ∣ ∣ ∣ \begin{matrix} - 2 sin (θ + α / 2) sin α / 2 & 2 sin α / 2 cos (θ + α / 2) - 2 sin α sin (θ + α) & 2 sin α cos (θ + α) \end{matrix} ∣ ∣ ∣$

$= 2 a b sin α sin \frac{α}{2} ∣ ∣ ∣ \begin{matrix} sin (θ + α / 2) & - cos (θ + α / 2) sin (θ + α) & - cos (θ + α) \end{matrix} ∣ ∣ ∣$

$= 2 a b sin α sin \frac{α}{2} sin \frac{α}{2}$
$= 2 a b {sin}^{2} \frac{α}{2} sin α$
$\Rightarrow Δ = 2 a b {sin}^{2} \frac{α}{2} sin α$
It is independent of $θ$

Now, $Δ = 2 a b {sin}^{2} \frac{α}{2} sin α$
$= a b (1 - cos α) sin α$
$= a b (sin α - sin α cos α)$
$\frac{d}{d α} (Δ) = a b (cos α - {cos}^{2} α + {sin}^{2} α)$
$\Rightarrow \frac{d}{d α} (Δ) = a b (cos α - cos 2 α)$
and $\frac{d^{2}}{d α^{2}} (Δ) = a b (- sin α + 2 sin 2 α)$

For $\frac{d}{d α} (Δ) = 0$
$cos α - cos 2 α = 0$
$\Rightarrow cos 2 α = cos (2 π - α)$
$\Rightarrow 2 α = 2 π - α$
$\Rightarrow α = \frac{2 π}{3}$

For $α = \frac{2 π}{3}$
$\frac{d^{2}}{d α^{2}} (Δ) = a b (- sin \frac{2 π}{3} + 2 sin \frac{4 π}{3}) < 0$
Therefore, area of $Δ A B C$ is maximum when $α = \frac{2 π}{3}$

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Similar questions

Q. Three points

A, B, C

are taken on an ellipse with eccentric angles

θ, θ + α

and

θ + 2 α

respectively. The area of

Δ A B C

Q. The points

P, Q, R

with eccentric angles

θ, θ + α, θ + 2 α

where

α \in (0, π)

are taken on the ellipse

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

then

Q. Consider three points

P = [- sin (β - α), - cos β], Q = [cos (β - α), sin β]

and

R = [cos (β - α + θ), sin (β - θ)],

where

0 < α, β, θ < \frac{π}{4} .

Then

Q. The area of the parallelogram formed by the tangents at the points whose eccentric angles are

θ, θ + \frac{π}{2}, θ + π, θ + \frac{3 π}{2}

on the ellipse

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

Two vectors having equal magnitudes of A make an angle θ with each other. Find the magnitude B of the resultant vector and the angle it makes with any one of the vectors?

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Three points A,B,C are taken on an ellipse with eccentric angles θ,θ+α and θ+2α respectively. The area of ΔABC is

Three points $A, B, C$ are taken on an ellipse with eccentric angles $θ, θ + α$ and $θ + 2 α$ respectively. The area of $Δ A B C$ is