To eliminate the parameter $t$ and express the relationship between $x$ and $y$ in Cartesian form, we’ll start with the given parametric equations:

$x = t^2 - 2$ $y = t + 3$

1. Solve for $t$ in terms of $y$ from the second equation: $y = t + 3 \Rightarrow t = y - 3$

2. Substitute $t = y - 3$ into the first equation to express $x$ in terms of $y$: $x = (y - 3)^2 - 2$

3. Expand and simplify: $x = (y^2 - 6y + 9) - 2$ $x = y^2 - 6y + 7$

So, the Cartesian equation of the curve is: $x = y^2 - 6y + 7$

Range of $y$:

Given that $-3 \leq t \leq 3$, substituting into $y = t + 3$ gives: $0 \leq y \leq 6$

Thus, the Cartesian equation is: $x = y^2 - 6y + 7 \quad \text{for} \quad 0 \leq y \leq 6$

Would you like more details on this or have any further questions?

Related Questions

1. How can we find the range of $x$ for this curve?
2. What is the vertex of the resulting parabola in terms of $x$ and $y$?
3. How would the equation change if the parametric equations included a different transformation?
4. Can we find the derivative $\frac{dy}{dx}$ of the curve given in Cartesian form?
5. How do we plot the Cartesian form of this parametric equation?

Tip

When eliminating parameters, isolate the parameter in one equation and substitute it into the other to find the direct relationship.