Correct Answer - Option 2 : 6/7

**Concept:**

**Slope of tangent to curve: **

- The first derivative of f(x).
- Plug x value of the indicated point into f '(x) to find the slope at x

**Calculation:**

\(\rm x=t^2+3t-8,\ y=2t^2-2t-5\)

\( \begin{aligned} &\rm \frac{d x}{d t}=2 t+3 \text { and } \frac{d y}{d t}=4 t-2\\\ &\rm \frac{\frac{d y}{dt}}{\frac{d x}{d t}}=\frac{4 t-2}{2 t+3}\\ &\rm\left.\frac{d y}{dx}\right|_{t=2}=\frac{4 (2)-2}{2 (2)+3}\\ &=\frac{6}{7} \end{aligned}\)

Hence, option (2) is correct.