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 solve the problem, we need to find the slope of the curve defined by the parametric equations \(x = t^2 + 3t - 8\) and \(y = 2t^2 - 2t - 5\) at the point \((2, -1)\).

### Step 1: Find the value of \(t\) at the point \((2, -1)\)

We start with the equation for \(x\):
\[ x = t^2 + 3t - 8 \]
Setting \(x = 2\):
\[ 2 = t^2 + 3t - 8 \]
Rearranging gives:
\[ t^2 + 3t - 10 = 0 \]

### Step 2: Factor the quadratic equation

We can factor the quadratic:
\[ t^2 + 5t - 2t - 10 = 0 \]
Grouping gives:
\[ t(t + 5) - 2(t + 5) = 0 \]
Factoring out \((t + 5)\):
\[ (t + 5)(t - 2) = 0 \]
Thus, the solutions for \(t\) are:
\[ t = -5 \quad \text{or} \quad t = 2 \]

### Step 3: Verify the value of \(t\) using \(y\)

Now we check which \(t\) value corresponds to \(y = -1\):
Using the equation for \(y\):
\[ y = 2t^2 - 2t - 5 \]
Setting \(y = -1\):
\[ -1 = 2t^2 - 2t - 5 \]
Rearranging gives:
\[ 2t^2 - 2t - 4 = 0 \]
Dividing by 2:
\[ t^2 - t - 2 = 0 \]

### Step 4: Factor the quadratic equation for \(y\)

Factoring gives:
\[ (t - 2)(t + 1) = 0 \]
Thus, the solutions for \(t\) are:
\[ t = 2 \quad \text{or} \quad t = -1 \]

### Step 5: Determine the valid \(t\)

From both equations, the common value of \(t\) is:
\[ t = 2 \]

### Step 6: Find \(\frac{dy}{dx}\)

To find the slope \(\frac{dy}{dx}\), we need to compute \(\frac{dy}{dt}\) and \(\frac{dx}{dt}\).

1. Differentiate \(x\) with respect to \(t\):
\[ \frac{dx}{dt} = 2t + 3 \]
2. Differentiate \(y\) with respect to \(t\):
\[ \frac{dy}{dt} = 4t - 2 \]

### Step 7: Compute \(\frac{dy}{dx}\)

Using the chain rule:
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{4t - 2}{2t + 3} \]

### Step 8: Evaluate \(\frac{dy}{dx}\) at \(t = 2\)

Substituting \(t = 2\):
\[ \frac{dy}{dx} = \frac{4(2) - 2}{2(2) + 3} = \frac{8 - 2}{4 + 3} = \frac{6}{7} \]

### Conclusion

The slope of the tangent to the curve at the point \((2, -1)\) is:
\[ \frac{dy}{dx} = \frac{6}{7} \]

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