To solve the problem step by step, we will follow these instructions:
### Step 1: Identify the equations
We have the line given by the equation:
\[ x + 2y = 1 \]
And the ellipse given by the equation:
\[ x^2 + 4y^2 = 1 \]
### Step 2: Find the points of intersection (A and B)
To find the points where the line intersects the ellipse, we will substitute \( y \) from the line equation into the ellipse equation.
From the line equation:
\[ y = \frac{1 - x}{2} \]
Substituting this into the ellipse equation:
\[
x^2 + 4\left(\frac{1 - x}{2}\right)^2 = 1
\]
\[
x^2 + 4\left(\frac{(1 - x)^2}{4}\right) = 1
\]
\[
x^2 + (1 - x)^2 = 1
\]
Expanding \( (1 - x)^2 \):
\[
x^2 + (1 - 2x + x^2) = 1
\]
\[
2x^2 - 2x + 1 = 1
\]
\[
2x^2 - 2x = 0
\]
Factoring out \( 2x \):
\[
2x(x - 1) = 0
\]
Thus, \( x = 0 \) or \( x = 1 \).
Now, substituting these values back into the line equation to find \( y \):
- For \( x = 0 \):
\[
0 + 2y = 1 \implies y = \frac{1}{2}
\]
So, point \( A(0, \frac{1}{2}) \).
- For \( x = 1 \):
\[
1 + 2y = 1 \implies 2y = 0 \implies y = 0
\]
So, point \( B(1, 0) \).
### Step 3: General point on the ellipse
Let point \( C \) on the ellipse be represented as:
\[ C(a, b) \]
where \( a = \cos \theta \) and \( b = \frac{1}{2} \sin \theta \) (since the ellipse can be parameterized as \( x = a \) and \( y = \frac{1}{2}b \)).
### Step 4: Area of triangle ABC
The area \( A \) of triangle \( ABC \) can be calculated using the formula:
\[
\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
\]
Substituting the coordinates of points \( A(0, \frac{1}{2}) \), \( B(1, 0) \), and \( C(\cos \theta, \frac{1}{2} \sin \theta) \):
\[
\text{Area} = \frac{1}{2} \left| 0(0 - \frac{1}{2} \sin \theta) + 1\left(\frac{1}{2} \sin \theta - \frac{1}{2}\right) + \cos \theta\left(\frac{1}{2} - 0\right) \right|
\]
\[
= \frac{1}{2} \left| \frac{1}{2} \sin \theta - \frac{1}{2} + \frac{1}{2} \cos \theta \right|
\]
\[
= \frac{1}{4} \left| \sin \theta + \cos \theta - 1 \right|
\]
### Step 5: Maximize the area
To maximize the area, we need to maximize \( \left| \sin \theta + \cos \theta - 1 \right| \).
Using the identity:
\[
\sin \theta + \cos \theta = \sqrt{2} \sin\left(\theta + \frac{\pi}{4}\right)
\]
The maximum value occurs when:
\[
\sin\left(\theta + \frac{\pi}{4}\right) = 1 \implies \theta + \frac{\pi}{4} = \frac{\pi}{2} \implies \theta = \frac{\pi}{4}
\]
### Step 6: Find coordinates of point C
Substituting \( \theta = \frac{\pi}{4} \):
\[
C\left(\cos\left(\frac{\pi}{4}\right), \frac{1}{2}\sin\left(\frac{\pi}{4}\right)\right) = C\left(\frac{1}{\sqrt{2}}, \frac{1}{2} \cdot \frac{1}{\sqrt{2}}\right) = C\left(\frac{1}{\sqrt{2}}, \frac{1}{2\sqrt{2}}\right)
\]
### Final Answer
The coordinates of point \( C \) are:
\[
C\left(\frac{1}{\sqrt{2}}, \frac{1}{2\sqrt{2}}\right)
\]
