For the curve `x=t^(2) +3t -8 ,y=2t^(2)-2t -5` at point (2,-1)

Options

A. length of subtangent is `7//6`
B. slope of tangent `=6//7`
C. length of tangent `=sqrt((85))//6`
D. none of these

Answer

To find the slope of the tangent to the curve defined by the parametric equations \( x = t^2 + 3t - 8 \) and \( y = 2t^2 - 2t - 5 \) at the point \( (2, -1) \), we will follow these steps:

### Step 1: Differentiate the parametric equations
We start by differentiating \( x \) and \( y \) with respect to \( t \).

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

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

### Step 2: Find the slope of the tangent
The slope of the tangent \( \frac{dy}{dx} \) can be found using the chain rule:
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{4t - 2}{2t + 3} \]

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

1. Set \( x = 2 \):
\[ t^2 + 3t - 8 = 2 \implies t^2 + 3t - 10 = 0 \]
Factor or use the quadratic formula:
\[ (t - 2)(t + 5) = 0 \implies t = 2 \text{ or } t = -5 \]

2. Set \( y = -1 \):
\[ 2t^2 - 2t - 5 = -1 \implies 2t^2 - 2t - 4 = 0 \implies t^2 - t - 2 = 0 \]
Factor or use the quadratic formula:
\[ (t - 2)(t + 1) = 0 \implies t = 2 \text{ or } t = -1 \]

### Step 4: Determine the common value of \( t \)
The common value of \( t \) that satisfies both equations is \( t = 2 \).

### Step 5: Calculate the slope at \( t = 2 \)
Substituting \( t = 2 \) into the slope formula:
\[ \frac{dy}{dx} = \frac{4(2) - 2}{2(2) + 3} = \frac{8 - 2}{4 + 3} = \frac{6}{7} \]

### Final Result
The slope of the tangent to the curve at the point \( (2, -1) \) is \( \frac{6}{7} \).

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