The equation of the normal to the curve parametrically represented by `x=t^(2)+3t-8 and y=2t^(2)-2t-5` at the point `P(2,-1)` is

Options

A. `2x+3y-1=0`
B. `6x-7y-11=0`
C. `7x+6y-8=0`
D. `3x+y-1=0`

Explanation

To find the equation of the normal to the curve represented parametrically by \( x = t^2 + 3t - 8 \) and \( y = 2t^2 - 2t - 5 \) at the point \( P(2, -1) \), we will follow these steps:

### Step 1: Differentiate the parametric equations
We need to find \( \frac{dy}{dt} \) and \( \frac{dx}{dt} \).

1. For \( x = t^2 + 3t - 8 \):
\[ \frac{dx}{dt} = 2t + 3 \]

2. For \( y = 2t^2 - 2t - 5 \):
\[ \frac{dy}{dt} = 4t - 2 \]

### Step 2: Find \( \frac{dy}{dx} \)
Using the chain rule, we can find \( \frac{dy}{dx} \) as follows:
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{4t - 2}{2t + 3} \]

### Step 3: Find the value of \( t \) at point \( P(2, -1) \)
We need to find \( t \) such that \( x(t) = 2 \) and \( y(t) = -1 \).

1. Set \( x = 2 \):
\[ 2 = t^2 + 3t - 8 \implies t^2 + 3t - 10 = 0 \]
Factoring gives:
\[ (t - 2)(t + 5) = 0 \implies t = 2 \text{ or } t = -5 \]

2. Set \( y = -1 \):
\[ -1 = 2t^2 - 2t - 5 \implies 2t^2 - 2t - 4 = 0 \implies t^2 - t - 2 = 0 \]
Factoring gives:
\[ (t - 2)(t + 1) = 0 \implies t = 2 \text{ or } t = -1 \]

Since \( t = 2 \) satisfies both equations, we will use \( t = 2 \).

### Step 4: Calculate the slope of the tangent line
Substituting \( t = 2 \) into \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{4(2) - 2}{2(2) + 3} = \frac{8 - 2}{4 + 3} = \frac{6}{7} \]

### Step 5: Calculate the slope of the normal line
The slope of the normal line is the negative reciprocal of the slope of the tangent line:
\[ \text{slope of normal} = -\frac{1}{\frac{6}{7}} = -\frac{7}{6} \]

### Step 6: Write the equation of the normal line
Using the point-slope form of the line equation:
\[ y - y_1 = m(x - x_1) \]
where \( (x_1, y_1) = (2, -1) \) and \( m = -\frac{7}{6} \):
\[ y + 1 = -\frac{7}{6}(x - 2) \]

### Step 7: Simplify the equation
Multiply through by 6 to eliminate the fraction:
\[ 6(y + 1) = -7(x - 2) \]
\[ 6y + 6 = -7x + 14 \]
Rearranging gives:
\[ 7x + 6y - 8 = 0 \]

Thus, the equation of the normal to the curve at the point \( P(2, -1) \) is:
\[ 7x + 6y - 8 = 0 \]