Equation of the normal of the curve $x=t^{2}$, $y=2t$ at $t=1$ is :

Question

Accepted Answer

To determine the equation of the normal to the curve given by the parametric equations $x = t^2$ and $y = 2t$ at $t = 1$, we follow these steps:

1. **Find the point on the curve at $t = 1$:**
   - When $t = 1$, $x = 1^2 = 1$ and $y = 2 \times 1 = 2$.
   - Therefore, the point on the curve is $(1, 2)$.

2. **Calculate the derivative $\frac{dy}{dx}$:**
   - From the parametric equations, $\frac{dx}{dt} = 2t$ and $\frac{dy}{dt} = 2$.
   - The derivative $\frac{dy}{dx}$ is given by $\frac{dy/dt}{dx/dt} = \frac{2}{2t} = \frac{1}{t}$.

3. **Evaluate the slope of the tangent at $t = 1$:**
   - At $t = 1$, the slope of the tangent is $\frac{1}{1} = 1$.

4. **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 $-1$.

5. **Write the equation of the normal line:**
   - The equation of a line with slope $m$ passing through a point $(x_1, y_1)$ is given by $y - y_1 = m(x - x_1)$.
   - Substituting the slope of the normal $-1$ and the point $(1, 2)$, we get:
     $$
     y - 2 = -1(x - 1)
     $$
     $$
     y - 2 = -x + 1
     $$
     $$
     x + y - 3 = 0
     $$

Thus, the equation of the normal is $x + y - 3 = 0$, which corresponds to Option 1.

**Explanation for why other options are incorrect:**

- **Option 2: $2x + y - 4 = 0$**
  - This equation implies a different slope and intercept, which does not match the calculated slope of $-1$ and the point $(1, 2)$.

- **Option 3: $x + y + 1 = 0$**
  - This equation suggests a different intercept, which does not satisfy the condition at the point $(1, 2)$.

- **Option 4: $x + y + 3 = 0$**
  - This equation also suggests a different intercept, which is inconsistent with the point $(1, 2)$.

Therefore, Option 1 is the correct answer as it accurately represents the equation of the normal to the curve at $t = 1$.