Find the equation of normal to the curves `x=t^2, y=2t+1` at point

Answer

To find the equation of the normal to the curve defined by the parametric equations \( x = t^2 \) and \( y = 2t + 1 \) at a point corresponding to the parameter \( t \), we can follow these steps:

### Step 1: Find the derivatives
Given the parametric equations:
\[ x = t^2 \]
\[ y = 2t + 1 \]

We differentiate \( x \) and \( y \) with respect to \( t \):
\[ \frac{dx}{dt} = 2t \]
\[ \frac{dy}{dt} = 2 \]

### Step 2: Find \(\frac{dy}{dx}\)
Using the chain rule, we find \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2}{2t} = \frac{1}{t} \]

### Step 3: Find the slope of the normal
The slope of the tangent line at the point is given by \(\frac{dy}{dx}\). Therefore, the slope of the normal line is:
\[ \text{slope of normal} = -\frac{1}{\frac{dy}{dx}} = -t \]

### Step 4: Find the coordinates of the point on the curve
At the parameter \( t \), the coordinates of the point on the curve are:
\[ (x, y) = (t^2, 2t + 1) \]

### Step 5: Write the equation of the normal line
Using the point-slope form of the equation of a line, the equation of the normal line at the point \((t^2, 2t + 1)\) with slope \(-t\) is:
\[ y - (2t + 1) = -t(x - t^2) \]

### Step 6: Simplify the equation
Expanding the equation:
\[ y - 2t - 1 = -tx + t^3 \]
Rearranging gives:
\[ y = -tx + t^3 + 2t + 1 \]

Thus, the equation of the normal to the curve at the point corresponding to the parameter \( t \) is:
\[ y = -tx + t^3 + 2t + 1 \]

### Summary
The equation of the normal to the curve \( x = t^2 \) and \( y = 2t + 1 \) at the point corresponding to the parameter \( t \) is:
\[ y = -tx + t^3 + 2t + 1 \]