A normal to $y=(x-1)^{2}$ at $x=2$ also passes through the point:

Question

Accepted Answer

To determine why Option 3 is the correct answer, we need to find the equation of the normal to the curve $ y = (x-1)^2 $ at the point where $ x = 2 $ and verify which of the given points it passes through.

1. **Find the point on the curve at $ x = 2 $:**

The curve is given by $ y = (x-1)^2 $. Substituting $ x = 2 $, we get:
   $$
   y = (2-1)^2 = 1^2 = 1
   $$
   So, the point on the curve is $ (2, 1) $.

2. **Calculate the derivative to find the slope of the tangent:**

The derivative of $ y = (x-1)^2 $ is:
   $$
   \frac{dy}{dx} = 2(x-1)
   $$
   At $ x = 2 $, the slope of the tangent is:
   $$
   2(2-1) = 2
   $$

3. **Determine the slope of the normal:**

The slope of the normal is the negative reciprocal of the slope of the tangent. Therefore, the slope of the normal is:
   $$
   -\frac{1}{2}
   $$

4. **Write the equation of the normal line:**

Using the point-slope form of a line equation, $ y - y_1 = m(x - x_1) $, where $ m $ is the slope and $ (x_1, y_1) $ is the point $ (2, 1) $:
   $$
   y - 1 = -\frac{1}{2}(x - 2)
   $$
   Simplifying, we get:
   $$
   y - 1 = -\frac{1}{2}x + 1
   $$
   $$
   y = -\frac{1}{2}x + 2
   $$

5. **Verify which point the normal passes through:**

We need to check which of the given points satisfies the equation $ y = -\frac{1}{2}x + 2 $.

- **Option 1: (0, 4)**
     $$
     4 \neq -\frac{1}{2}(0) + 2 = 2
     $$
     This point does not lie on the line.

- **Option 2: (0, 0)**
     $$
     0 \neq -\frac{1}{2}(0) + 2 = 2
     $$
     This point does not lie on the line.

- **Option 3: (4, 0)**
     $$
     0 = -\frac{1}{2}(4) + 2 = -2 + 2 = 0
     $$
     This point satisfies the equation of the normal line.

- **Option 4: (1, 2)**
     $$
     2 \neq -\frac{1}{2}(1) + 2 = -\frac{1}{2} + 2 = \frac{3}{2}
     $$
     This point does not lie on the line.

Therefore, the correct answer is Option 3: (4, 0), as it is the only point that lies on the normal line to the curve at $ x = 2 $.