3.4 Parametric equations and Cartesian conversion — Practise

Work through these on your own. Before you eliminate the parameter, choose your route: substitute from the simpler equation, or square and add with $\cos^2 t + \sin^2 t = 1$ . When the parameter has a restricted range, carry that range across to the Cartesian form. Later items reach back to simultaneous equations and to circles from 3.3, so decide which tool each one needs first.

Both the questions and the answers below come from a set that is checked by computer every time the book is built.

A curve has parametric equations $x = 3t$ , $y = t^2 + 1$ . Find the point where $t = 2$ .
Find the Cartesian equation of the curve $x = 3t$ , $y = t^2 + 1$ .
Find the Cartesian equation of the curve $x = t - 4$ , $y = 2t + 1$ .
Find the Cartesian equation of the curve $x = 4\cos t$ , $y = 4\sin t$ .
Find the Cartesian equation of the curve $x = 1 + 3\cos t$ , $y = -2 + 3\sin t$ .
Give parametric equations for the curve $y = x^2 - 3$ , and show that eliminating the parameter returns the same curve.
A curve is given by $x = t + 2$ , $y = t^2$ , for $0 \le t \le 2$ . Find its Cartesian equation and the range of $x$ that it covers.
Find the Cartesian equation of the curve $x = t$ , $y = \dfrac{1}{t}$ , where $t \ne 0$ .
A curve is given by $x = t^2$ , $y = t^3$ . Show that it satisfies $y^2 = x^3$ .
Decide whether the point $(4, 8)$ lies on the curve $x = t^2$ , $y = t^3$ .
Find the Cartesian equation of the curve $x = 2t + 1$ , $y = t - 3$ .
A curve is given by $x = t^2$ , $y = 2t$ . Find the Cartesian equation, and state the part of the curve it covers.

Answers and concise solutions

1. The point

Answer: Substitute $t = 2$ into each equation: $x = 3 \times 2 = 6$ and $y = 2^2 + 1 = 5$ . The point is $(6, 5)$ .

2. The Cartesian equation

Answer: Make $t$ the subject of $x = 3t$ : $t = \dfrac{x}{3}$ . Substitute into $y = t^2 + 1$ : $y = \left(\dfrac{x}{3}\right)^2 + 1 = \dfrac{x^2}{9} + 1$ . The curve is a parabola. Check the point from the previous question: at $x = 6$ , $y = \frac{36}{9} + 1 = 5$ .

3. The Cartesian equation

Answer: Make $t$ the subject of $x = t - 4$ : $t = x + 4$ . Substitute into $y = 2t + 1$ : $y = 2(x + 4) + 1 = 2x + 9$ . The curve is a straight line of gradient $2$ .

4. The Cartesian equation

Answer: Make each trigonometric function the subject: $\cos t = \dfrac{x}{4}$ and $\sin t = \dfrac{y}{4}$ . Square and add, using $\cos^2 t + \sin^2 t = 1$ : $\left(\dfrac{x}{4}\right)^2 + \left(\dfrac{y}{4}\right)^2 = 1$ . Multiply by $16$ to get $x^2 + y^2 = 16$ , the circle of radius $4$ centred on the origin.

5. The Cartesian equation

Answer: Rearrange first: $\cos t = \dfrac{x - 1}{3}$ and $\sin t = \dfrac{y + 2}{3}$ . Square and add: $\left(\dfrac{x - 1}{3}\right)^2 + \left(\dfrac{y + 2}{3}\right)^2 = 1$ , so $(x - 1)^2 + (y + 2)^2 = 9$ . The curve is a circle with centre $(1, -2)$ and radius $3$ .

6. The parametric equations

Answer: Let $t = x$ . Then $x = t$ and $y = t^2 - 3$ . Eliminating $t$ by substituting $t = x$ gives back $y = x^2 - 3$ , so the parameterisation describes the whole curve. This choice of parameter is the simplest one; a curve has many parameterisations.

7. The Cartesian equation and the range of x

Answer: Eliminate $t$ : from $x = t + 2$ we get $t = x - 2$ , so $y = (x - 2)^2$ . The parameter runs from $0$ to $2$ , so $x = t + 2$ runs from $2$ to $4$ . The curve is the part of the parabola with $2 \le x \le 4$ , not the whole parabola.

8. The Cartesian equation

Answer: The first equation already gives $t = x$ . Substitute into the second: $y = \dfrac{1}{x}$ . The curve is a hyperbola. The condition $t \ne 0$ becomes $x \ne 0$ , so the curve is not defined at the origin and neither axis is crossed.

9. The identity

Answer: Substitute both parametric expressions: $y^2 = (t^3)^2 = t^6$ and $x^3 = (t^2)^3 = t^6$ . These are equal, so every point of the curve satisfies $y^2 = x^3$ . Eliminating $t$ directly is not possible here, because $x = t^2$ gives two values of $t$ , $t = \sqrt{x}$ and $t = -\sqrt{x}$ ; the curve is not the graph of a function.

10. The decision

Answer: Try $t = 2$ : $x = 2^2 = 4$ and $y = 2^3 = 8$ . Both coordinates match, so $(4, 8)$ lies on the curve. The parameter $t = 2$ produces the point, which is what "lies on the curve" means for a parametric description.

11. The Cartesian equation

Answer: Either equation can give $t$ . From $x = 2t + 1$ : $t = \dfrac{x - 1}{2}$ . Substitute: $y = \dfrac{x - 1}{2} - 3 = \dfrac{x}{2} - \dfrac{1}{2} - 3 = \dfrac{x}{2} - \dfrac{7}{2}$ . The curve is a straight line. Check with the other equation: from $y = t - 3$ , $t = y + 3$ ; substituting into $x = 2t + 1$ gives $x = 2y + 7$ . Rearranging gives $y = \frac{x}{2} - \frac{7}{2}$ , the same line.

12. The Cartesian equation and the range of x

Answer: Eliminate $t$ from the equation without a square: $t = \dfrac{y}{2}$ , so $x = \left(\dfrac{y}{2}\right)^2 = \dfrac{y^2}{4}$ . This is the same as $y^2 = 4x$ , a parabola opening to the right. Every value of $x = t^2$ is at least $0$ , so the curve covers only $x \ge 0$ . Eliminating the parameter with the squared equation would give two branches and lose this restriction.