Answer :
Let's solve the problem step by step.
Given:
The curve is [tex]\( 2y = 3x^3 - 7x^2 + 4x \)[/tex].
First, we rewrite the curve equation for simplicity:
[tex]\[ y = \frac{1}{2} (3x^3 - 7x^2 + 4x) \][/tex]
To find the normals at points [tex]\( O(0, 0) \)[/tex] and [tex]\( A(1, 0) \)[/tex], we first need the derivative of [tex]\( y \)[/tex] with respect to [tex]\( x \)[/tex].
### Step 1: Find the derivative of [tex]\( y \)[/tex]
[tex]\[ y = \frac{1}{2} (3x^3 - 7x^2 + 4x) \][/tex]
[tex]\[ y = \frac{1}{2} (3x^3) - \frac{1}{2} (7x^2) + \frac{1}{2} (4x) \][/tex]
[tex]\[ y = \frac{3}{2}x^3 - \frac{7}{2}x^2 + 2x \][/tex]
Now, differentiate [tex]\( y \)[/tex] with respect to [tex]\( x \)[/tex]:
[tex]\[ \frac{dy}{dx} = \frac{d}{dx} \left( \frac{3}{2}x^3 - \frac{7}{2}x^2 + 2x \right) \][/tex]
[tex]\[ \frac{dy}{dx} = \frac{3}{2} \cdot 3x^2 - \frac{7}{2} \cdot 2x + 2 \][/tex]
[tex]\[ \frac{dy}{dx} = \frac{9}{2}x^2 - 7x + 2 \][/tex]
### Step 2: Find the slopes of the tangent lines at points [tex]\( O \)[/tex] and [tex]\( A \)[/tex]
For point [tex]\( O(0, 0) \)[/tex]:
[tex]\[ \left. \frac{dy}{dx} \right|_{x=0} = \frac{9}{2} \cdot 0^2 - 7 \cdot 0 + 2 = 2 \][/tex]
For point [tex]\( A(1, 0) \)[/tex]:
[tex]\[ \left. \frac{dy}{dx} \right|_{x=1} = \frac{9}{2} \cdot 1^2 - 7 \cdot 1 + 2 \][/tex]
[tex]\[ \left. \frac{dy}{dx} \right|_{x=1} = \frac{9}{2} - 7 + 2 \][/tex]
[tex]\[ \left. \frac{dy}{dx} \right|_{x=1} = \frac{9}{2} - \frac{14}{2} + \frac{4}{2} \][/tex]
[tex]\[ \left. \frac{dy}{dx} \right|_{x=1} = \frac{-1}{2} \][/tex]
### Step 3: Find the slopes of the normals
The slope of the normal is the negative reciprocal of the slope of the tangent.
For the normal at [tex]\( O(0, 0) \)[/tex]:
[tex]\[ \text{Slope of normal} = -\frac{1}{2} \][/tex]
For the normal at [tex]\( A(1, 0) \)[/tex]:
[tex]\[ \text{Slope of normal} = -\left( \frac{1}{-\frac{1}{2}} \right) = 2 \][/tex]
### Step 4: Find the equations of the normals
For the normal at [tex]\( O(0, 0) \)[/tex]:
[tex]\[ y = -\frac{1}{2} x \][/tex]
For the normal at [tex]\( A(1, 0) \)[/tex]:
Using the point-slope form of a line equation, [tex]\( y - y_1 = m(x - x_1) \)[/tex]:
[tex]\[ y - 0 = 2(x - 1) \][/tex]
[tex]\[ y = 2x - 2 \][/tex]
### Step 5: Solve for the intersection of the normals to find [tex]\( N \)[/tex]
Set the equations equal to each other:
[tex]\[ -\frac{1}{2} x = 2x - 2 \][/tex]
Multiply by 2 to clear the fraction:
[tex]\[ -x = 4x - 4 \][/tex]
[tex]\[ -x - 4x = -4 \][/tex]
[tex]\[ -5x = -4 \][/tex]
[tex]\[ x = \frac{4}{5} \][/tex]
Substitute [tex]\( x = \frac{4}{5} \)[/tex] into one of the normal equations to find [tex]\( y \)[/tex]:
[tex]\[ y = -\frac{1}{2} \left( \frac{4}{5} \right) \][/tex]
[tex]\[ y = -\frac{2}{5} \][/tex]
Thus, the coordinates of [tex]\( N \)[/tex] are [tex]\( \left( \frac{4}{5}, -\frac{2}{5} \right) \)[/tex].
### Step 6: Calculate the area of triangle [tex]\( OAN \)[/tex]
The vertices of the triangle [tex]\( OAN \)[/tex] are [tex]\( O(0, 0) \)[/tex], [tex]\( A(1, 0) \)[/tex], and [tex]\( N \left( \frac{4}{5}, -\frac{2}{5} \right) \)[/tex].
The area of a triangle with vertices [tex]\((x_1, y_1)\)[/tex], [tex]\((x_2, y_2)\)[/tex], [tex]\((x_3, y_3)\)[/tex] is given by:
[tex]\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \][/tex]
Substitute the coordinates:
[tex]\[ \text{Area} = \frac{1}{2} \left| 0(0 - \left( -\frac{2}{5} \right)) + 1\left( -\frac{2}{5} - 0 \right) + \frac{4}{5}(0 - 0) \right| \][/tex]
[tex]\[ \text{Area} = \frac{1}{2} \left| 0 + \left( -\frac{2}{5} \right) + 0 \right| \][/tex]
[tex]\[ \text{Area} = \frac{1}{2} \left| -\frac{2}{5} \right| \][/tex]
[tex]\[ \text{Area} = \frac{1}{2} \times \frac{2}{5} \][/tex]
[tex]\[ \text{Area} = \frac{1}{5} \][/tex]
Hence, the area of triangle [tex]\( OAN \)[/tex] is [tex]\( \frac{1}{5} \)[/tex] square units.
### Summary
a) The coordinates of [tex]\( N \)[/tex] are [tex]\( \left( \frac{4}{5}, -\frac{2}{5} \right) \)[/tex].
b) The area of triangle [tex]\( OAN \)[/tex] is [tex]\( \frac{1}{5} \)[/tex] square units.
Given:
The curve is [tex]\( 2y = 3x^3 - 7x^2 + 4x \)[/tex].
First, we rewrite the curve equation for simplicity:
[tex]\[ y = \frac{1}{2} (3x^3 - 7x^2 + 4x) \][/tex]
To find the normals at points [tex]\( O(0, 0) \)[/tex] and [tex]\( A(1, 0) \)[/tex], we first need the derivative of [tex]\( y \)[/tex] with respect to [tex]\( x \)[/tex].
### Step 1: Find the derivative of [tex]\( y \)[/tex]
[tex]\[ y = \frac{1}{2} (3x^3 - 7x^2 + 4x) \][/tex]
[tex]\[ y = \frac{1}{2} (3x^3) - \frac{1}{2} (7x^2) + \frac{1}{2} (4x) \][/tex]
[tex]\[ y = \frac{3}{2}x^3 - \frac{7}{2}x^2 + 2x \][/tex]
Now, differentiate [tex]\( y \)[/tex] with respect to [tex]\( x \)[/tex]:
[tex]\[ \frac{dy}{dx} = \frac{d}{dx} \left( \frac{3}{2}x^3 - \frac{7}{2}x^2 + 2x \right) \][/tex]
[tex]\[ \frac{dy}{dx} = \frac{3}{2} \cdot 3x^2 - \frac{7}{2} \cdot 2x + 2 \][/tex]
[tex]\[ \frac{dy}{dx} = \frac{9}{2}x^2 - 7x + 2 \][/tex]
### Step 2: Find the slopes of the tangent lines at points [tex]\( O \)[/tex] and [tex]\( A \)[/tex]
For point [tex]\( O(0, 0) \)[/tex]:
[tex]\[ \left. \frac{dy}{dx} \right|_{x=0} = \frac{9}{2} \cdot 0^2 - 7 \cdot 0 + 2 = 2 \][/tex]
For point [tex]\( A(1, 0) \)[/tex]:
[tex]\[ \left. \frac{dy}{dx} \right|_{x=1} = \frac{9}{2} \cdot 1^2 - 7 \cdot 1 + 2 \][/tex]
[tex]\[ \left. \frac{dy}{dx} \right|_{x=1} = \frac{9}{2} - 7 + 2 \][/tex]
[tex]\[ \left. \frac{dy}{dx} \right|_{x=1} = \frac{9}{2} - \frac{14}{2} + \frac{4}{2} \][/tex]
[tex]\[ \left. \frac{dy}{dx} \right|_{x=1} = \frac{-1}{2} \][/tex]
### Step 3: Find the slopes of the normals
The slope of the normal is the negative reciprocal of the slope of the tangent.
For the normal at [tex]\( O(0, 0) \)[/tex]:
[tex]\[ \text{Slope of normal} = -\frac{1}{2} \][/tex]
For the normal at [tex]\( A(1, 0) \)[/tex]:
[tex]\[ \text{Slope of normal} = -\left( \frac{1}{-\frac{1}{2}} \right) = 2 \][/tex]
### Step 4: Find the equations of the normals
For the normal at [tex]\( O(0, 0) \)[/tex]:
[tex]\[ y = -\frac{1}{2} x \][/tex]
For the normal at [tex]\( A(1, 0) \)[/tex]:
Using the point-slope form of a line equation, [tex]\( y - y_1 = m(x - x_1) \)[/tex]:
[tex]\[ y - 0 = 2(x - 1) \][/tex]
[tex]\[ y = 2x - 2 \][/tex]
### Step 5: Solve for the intersection of the normals to find [tex]\( N \)[/tex]
Set the equations equal to each other:
[tex]\[ -\frac{1}{2} x = 2x - 2 \][/tex]
Multiply by 2 to clear the fraction:
[tex]\[ -x = 4x - 4 \][/tex]
[tex]\[ -x - 4x = -4 \][/tex]
[tex]\[ -5x = -4 \][/tex]
[tex]\[ x = \frac{4}{5} \][/tex]
Substitute [tex]\( x = \frac{4}{5} \)[/tex] into one of the normal equations to find [tex]\( y \)[/tex]:
[tex]\[ y = -\frac{1}{2} \left( \frac{4}{5} \right) \][/tex]
[tex]\[ y = -\frac{2}{5} \][/tex]
Thus, the coordinates of [tex]\( N \)[/tex] are [tex]\( \left( \frac{4}{5}, -\frac{2}{5} \right) \)[/tex].
### Step 6: Calculate the area of triangle [tex]\( OAN \)[/tex]
The vertices of the triangle [tex]\( OAN \)[/tex] are [tex]\( O(0, 0) \)[/tex], [tex]\( A(1, 0) \)[/tex], and [tex]\( N \left( \frac{4}{5}, -\frac{2}{5} \right) \)[/tex].
The area of a triangle with vertices [tex]\((x_1, y_1)\)[/tex], [tex]\((x_2, y_2)\)[/tex], [tex]\((x_3, y_3)\)[/tex] is given by:
[tex]\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \][/tex]
Substitute the coordinates:
[tex]\[ \text{Area} = \frac{1}{2} \left| 0(0 - \left( -\frac{2}{5} \right)) + 1\left( -\frac{2}{5} - 0 \right) + \frac{4}{5}(0 - 0) \right| \][/tex]
[tex]\[ \text{Area} = \frac{1}{2} \left| 0 + \left( -\frac{2}{5} \right) + 0 \right| \][/tex]
[tex]\[ \text{Area} = \frac{1}{2} \left| -\frac{2}{5} \right| \][/tex]
[tex]\[ \text{Area} = \frac{1}{2} \times \frac{2}{5} \][/tex]
[tex]\[ \text{Area} = \frac{1}{5} \][/tex]
Hence, the area of triangle [tex]\( OAN \)[/tex] is [tex]\( \frac{1}{5} \)[/tex] square units.
### Summary
a) The coordinates of [tex]\( N \)[/tex] are [tex]\( \left( \frac{4}{5}, -\frac{2}{5} \right) \)[/tex].
b) The area of triangle [tex]\( OAN \)[/tex] is [tex]\( \frac{1}{5} \)[/tex] square units.
