Two points A(1, 4) & B(3, 0) are given on the ellipse `2x^(2) + y^(2) = 18`. The co-ordinates of a point C on the ellipse such that the area of the triangle ABC is greatest is

Options

A. `(sqrt6 , sqrt6)`
B. `(-sqrt6 , sqrt6)`
C. `(sqrt6 , -sqrt6)`
D. `(-sqrt6 , -sqrt6)`

Answer

To find the coordinates of point C on the ellipse \(2x^2 + y^2 = 18\) such that the area of triangle ABC is maximized, we can follow these steps:

### Step 1: Understand the Area of Triangle Formula
The area \(A\) of triangle formed by points \(A(x_1, y_1)\), \(B(x_2, y_2)\), and \(C(x_3, y_3)\) can be calculated using the formula:

\[ A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]

### Step 2: Define the Points
Given points are:
- \(A(1, 4)\)
- \(B(3, 0)\)
- Let \(C\) be any point on the ellipse, represented as \(C(x, y)\).

### Step 3: Express \(y\) in terms of \(x\)
From the equation of the ellipse \(2x^2 + y^2 = 18\), we can express \(y\) in terms of \(x\):

\[ y = \sqrt{18 - 2x^2} \]

### Step 4: Substitute Coordinates into Area Formula
Substituting the coordinates of points A, B, and C into the area formula:

\[ A = \frac{1}{2} \left| 1(0 - y) + 3(y - 4) + x(4 - 0) \right| \]

This simplifies to:

\[ A = \frac{1}{2} \left| -y + 3y - 12 + 4x \right| \]
\[ A = \frac{1}{2} \left| 2y + 4x - 12 \right| \]

### Step 5: Substitute \(y\) into the Area Expression
Substituting \(y = \sqrt{18 - 2x^2}\):

\[ A = \frac{1}{2} \left| 2\sqrt{18 - 2x^2} + 4x - 12 \right| \]

### Step 6: Differentiate the Area with Respect to \(x\)
To find the maximum area, we need to differentiate \(A\) with respect to \(x\) and set the derivative to zero.

Let \(A = \frac{1}{2} \left( 2\sqrt{18 - 2x^2} + 4x - 12 \right)\).

Differentiating \(A\):

\[ \frac{dA}{dx} = \frac{1}{2} \left( \frac{-2x}{\sqrt{18 - 2x^2}} + 4 \right) \]

Setting \(\frac{dA}{dx} = 0\):

\[ \frac{-2x}{\sqrt{18 - 2x^2}} + 4 = 0 \]

### Step 7: Solve for \(x\)
Rearranging gives:

\[ \frac{-2x}{\sqrt{18 - 2x^2}} = -4 \]

Cross-multiplying:

\[ 2x = 4\sqrt{18 - 2x^2} \]

Squaring both sides:

\[ 4x^2 = 16(18 - 2x^2) \]

Expanding and rearranging:

\[ 4x^2 + 32x^2 - 288 = 0 \]
\[ 36x^2 = 288 \]
\[ x^2 = 8 \quad \Rightarrow \quad x = \pm 2\sqrt{2} \]

### Step 8: Find Corresponding \(y\)
Substituting \(x = 2\sqrt{2}\) back into the ellipse equation to find \(y\):

\[ y = \sqrt{18 - 2(2\sqrt{2})^2} = \sqrt{18 - 8} = \sqrt{10} \]

### Final Coordinates of Point C
Thus, the coordinates of point C that maximize the area of triangle ABC are:

\[ C(2\sqrt{2}, \sqrt{10}) \]