Find the equation of normal of the curve `2y= 7x - 5x^(2)` at those points at which the curve intersects the line x = y.

Answer

To find the equation of the normal to the curve \( 2y = 7x - 5x^2 \) at the points where the curve intersects the line \( x = y \), we will follow these steps:

### Step 1: Find the points of intersection of the curve and the line
We start by substituting \( y = x \) into the curve equation:

\[ 2y = 7x - 5x^2 \implies 2x = 7x - 5x^2 \]

Rearranging gives:

\[ 5x^2 - 5x = 0 \]

Factoring out \( 5x \):

\[ 5x(x - 1) = 0 \]

Thus, we find:

\[ x = 0 \quad \text{or} \quad x = 1 \]

Now, substituting back to find \( y \):

- For \( x = 0 \): \( y = 0 \)
- For \( x = 1 \): \( y = 1 \)

The points of intersection are \( (0, 0) \) and \( (1, 1) \).

### Step 2: Find the slope of the tangent to the curve
We differentiate the curve equation \( 2y = 7x - 5x^2 \) with respect to \( x \):

\[ \frac{d}{dx}(2y) = \frac{d}{dx}(7x - 5x^2) \implies 2\frac{dy}{dx} = 7 - 10x \]

Thus, we have:

\[ \frac{dy}{dx} = \frac{7 - 10x}{2} \]

### Step 3: Find the slope of the normal
The slope of the normal is the negative reciprocal of the slope of the tangent. Therefore:

\[ \text{slope of normal} = -\frac{1}{\frac{dy}{dx}} = -\frac{2}{7 - 10x} \]

### Step 4: Find the equation of the normal at each point of intersection

**At point \( (0, 0) \)**:

1. Calculate the slope of the normal:
\[ \text{slope at } x = 0 = -\frac{2}{7 - 10(0)} = -\frac{2}{7} \]

2. Use the point-slope form of the equation of the normal:
\[ y - y_1 = m(x - x_1) \implies y - 0 = -\frac{2}{7}(x - 0) \]
Simplifying gives:
\[ y = -\frac{2}{7}x \]
Rearranging:
\[ 2x + 7y = 0 \]

**At point \( (1, 1) \)**:

1. Calculate the slope of the normal:
\[ \text{slope at } x = 1 = -\frac{2}{7 - 10(1)} = -\frac{2}{-3} = \frac{2}{3} \]

2. Use the point-slope form of the equation of the normal:
\[ y - 1 = \frac{2}{3}(x - 1) \]
Simplifying gives:
\[ y - 1 = \frac{2}{3}x - \frac{2}{3} \implies y = \frac{2}{3}x + \frac{1}{3} \]
Rearranging:
\[ 2x - 3y + 1 = 0 \]

### Final Results
The equations of the normals at the points of intersection are:

1. At \( (0, 0) \): \( 2x + 7y = 0 \)
2. At \( (1, 1) \): \( 2x - 3y + 1 = 0 \)