Precalculus

Parametric equations

An equation in x and y tells you where a path lies. How do you also say where a moving object is at each moment, and which way it is going?

You need

Magnitude and direction of a vectorPrecalculus VectorsGrades 11 to 12 Trig functions on the unit circlePrecalculus Function notationGrades 8 to 9 What a function isGrades 8 to 9 +2 more

Unlocks

Eliminating the parameterPrecalculus Parametric, polar and vector valued functionsGrade 12

Where you are 10 in Polar coordinates and vectors

Trigonometry7

Polar coordinates and vectors10

Parametric equations

Polar coordinates3 Vector addition and scalar multiplication3 Vectors4 Vectors in space ▾ Show all 10 (5 more)

Why this exists

The equation (x - 3)^2 + (y + 1)^2 = 4 describes a circle all at once. It cannot say where a point on the circle is after 5 seconds, where it started, or which way it goes around. It is not even a function of x, since a vertical line can cross a circle twice (see What a function is).

A moving object raises exactly these questions. A robot mower crossing a lawn or a ball flying through the air has an x-coordinate that changes with time and a y-coordinate that changes with time. The natural description is two functions of time, one for each coordinate. That pair carries what a single equation in x and y throws away: the order in which the points are visited.

The lesson takes two days. Part 1 defines parametric equations, plots them from a table, reads their direction and writes a line through two points. Part 2 traces circles and ellipses with cosine and sine, and models a thrown ball.

Definition

Let t run through an interval I of real numbers. Two functions x = f(t) and y = g(t) on I are parametric equations, and t is the parameter. Each value of t gives one point (x(t), y(t)), and the set of all these points is a plane curve.

The orientation of the curve is the direction in which its points are traced as t increases, shown by arrows along the curve. When I is a \le t \le b, the point at t = a is the initial point and the point at t = b is the terminal point. The parameter is often time, but it need not be; it is the input both coordinates depend on.

Intuition

Think of t as a clock. At each tick, f says how far across the point is and g says how high. The curve is the trail the point leaves, and the arrows record which way it was moving.

Figure 1One circle, two parametrizations. The parameter decides where the trip starts and which way it goes.

The trail alone does not fix the motion. In Figure 1 both pairs draw the same circle of radius 2. On the left, x = 2\cos t, y = 2\sin t starts on the right and goes counterclockwise. On the right, x = 2\sin t, y = 2\cos t starts at the top and goes clockwise. Same picture, different trips: parametric equations add a starting point, a direction and a pace.

Results

Part 1: a line. For a point (x_0, y_0) and numbers a and b, not both zero,

x = x_0 + at, \qquad y = y_0 + bt

trace a straight line. The point is at (x_0, y_0) when t = 0 and moves a across and b up per unit of t, a distance of \displaystyle \sqrt{a^2 + b^2}. If a \ne 0 the slope is \displaystyle \frac{b}{a}. To go from (x_0, y_0) to (x_1, y_1) as t runs from 0 to T, take \displaystyle a = \frac{x_1 - x_0}{T} and \displaystyle b = \frac{y_1 - y_0}{T}.

Part 2: a circle. For a center (h, k) and radius r > 0,

x = h + r\cos t, \qquad y = k + r\sin t

with 0 \le t \le 2\pi trace the circle once counterclockwise, starting at the rightmost point (h + r, k). Replacing t by \omega t with \omega > 0 keeps the circle and direction, and one turn takes \displaystyle \frac{2\pi}{\omega} units of t. Using y = k - r\sin t instead reverses the direction to clockwise.

Part 2: an ellipse. With a, b > 0, the equations x = h + a\cos t, y = k + b\sin t, 0 \le t \le 2\pi, trace once counterclockwise the ellipse centered at (h, k) that reaches a units left and right of the center and b units up and down.

Part 2: a projectile. A ball launched at t = 0 from height h, with speed v_0 at angle \theta above the horizontal, is modeled by

\begin{aligned} x &= (v_0\cos\theta)\,t,\\ y &= h + (v_0\sin\theta)\,t - 16t^2, \end{aligned}

with distances in feet and t in seconds. In meters the last term is 4.9t^2. The model assumes no air resistance, a steady downward pull of gravity, and flat ground at height 0. It holds from launch until the ball lands or hits something.

Why it is true

The line. Each coordinate is a linear function of t, so between any two times the point moves a\,\Delta t across and b\,\Delta t up. The ratio of rise to run is always \displaystyle \frac{b}{a}, and a path with the same slope between any two of its points is a straight line. In vector words, the position is the start \langle x_0, y_0 \rangle plus t copies of \langle a, b \rangle, and by Magnitude and direction of a vector each copy has length \displaystyle \sqrt{a^2 + b^2}.

The circle. By Trig functions on the unit circle, (\cos t, \sin t) is the terminal point of the angle t on the unit circle. Multiplying by r gives the point at distance r from the origin in the same direction, and adding (h, k) slides it to distance r from (h, k). A direct check:

(x - h)^2 + (y - k)^2 = r^2\cos^2 t + r^2\sin^2 t = r^2.

So every point is on the circle with center (h, k) and radius r (see Equation of a circle). As t grows from 0 to 2\pi the angle sweeps once counterclockwise, so the point goes once around. With y = k - r\sin t the point is the mirror image across the line y = k, and a mirror reverses the direction of turning. The ellipse is the same argument with different stretches a and b.

The projectile. The launch velocity is a vector of magnitude v_0 at angle \theta, so its components are v_0\cos\theta across and v_0\sin\theta up. With no air resistance nothing pushes the ball sideways, so x grows at the steady rate v_0\cos\theta. Vertically, gravity pulls the ball below the straight line h + (v_0\sin\theta)t by 16t^2 feet after t seconds, the height law used in Maximum, minimum and projectile problems.

Worked examples

Example 1

Part 1: plotting from a table

Plot the curve x = t^2 - 2t, y = t + 1 for -1 \le t \le 3. Give the initial and terminal points, the orientation, and the leftmost point.

Solution

Each value of t gives one point, so make a table, working out x and y from the same t in each row.

t	x = t^2 - 2t	y = t + 1
-1	3	0
0	0	1
1	-1	2
2	0	3
3	3	4

Plot the points and join them in order of t, not of x. The initial point is (3, 0) and the terminal point is (3, 4). Since y = t + 1 increases with t, the point always moves upward, so the arrows point up along the curve: it swings left and back right, a parabola opening to the right.

Figure 2Example 1. The table gives the dots; the order of the dots gives the arrows.

For the leftmost point, complete the square: x = (t - 1)^2 - 1. A square is never negative, so x \ge -1, with x = -1 only at t = 1. The leftmost point is (-1, 2).

Example 2

Part 1: a mower between two points

Ngozika’s robot mower drives straight across a lawn at a steady pace, from (2, -3) to (14, 2) in 5 seconds, with coordinates in meters. Write its position t seconds after it starts, find where it is at t = 2, and find its speed.

Solution

In 5 seconds x changes by 14 - 2 = 12 meters, a steady 2.4 meters per second, and y changes by 2 - (-3) = 5 meters, or 1 meter per second. Starting from (2, -3):

x = 2 + 2.4t, \qquad y = -3 + t

for 0 \le t \le 5. Check the end: at t = 5 the point is (14, 2). At t = 2, x = 2 + 4.8 and y = -1, so the mower is at (6.8, -1).

Each second it moves along \langle 2.4, 1 \rangle, whose magnitude is \displaystyle \sqrt{2.4^2 + 1^2} = \sqrt{6.76} = 2.6. Its speed is 2.6 meters per second. As a check, the whole trip is \displaystyle \sqrt{12^2 + 5^2} = 13 meters in 5 seconds.

Example 3

Part 1: the same segment, the other way

On the return trip the mower follows x = 14 - 6t, y = 2 - 2.5t for 0 \le t \le 2. Show that it covers the segment of Example 2, and compare the direction, the speed, and when each trip passes (8, -0.5).

Solution

At t = 0 the mower is at (14, 2) and at t = 2 at (2, -3): the same ends, swapped. To see that every point is shared, put s = 5 - 2.5t into Example 2:

\begin{aligned} 2 + 2.4(5 - 2.5t) &= 14 - 6t,\\ -3 + (5 - 2.5t) &= 2 - 2.5t. \end{aligned}

So the return trip at time t is where the first trip was at time s, and as t runs from 0 to 2, s runs from 5 down to 0. Same segment, opposite orientation.

The speed is \displaystyle \sqrt{(-6)^2 + (-2.5)^2} = \sqrt{42.25} = 6.5 meters per second, two and a half times as fast. The first trip has x = 8 when 2 + 2.4t = 8, so t = 2.5, and then y = -0.5. The return trip has x = 8 when 14 - 6t = 8, so t = 1, and then y = -0.5. One point, visited at different times: the curve is the same, the parametrizations are not.

Example 4

Part 2: a circle from its equations

Describe the curve x = 3 + 2\cos t, y = -1 + 2\sin t for 0 \le t \le 2\pi. Then say what changes on 0 \le t \le \pi, and what changes if y = -1 - 2\sin t.

Solution

This is the circle form with h = 3, k = -1 and r = 2. The quarter turns confirm it:

t	(x, y)
0	(5, -1)
\displaystyle \frac{\pi}{2}	(3, 1)
\pi	(1, -1)
\displaystyle \frac{3\pi}{2}	(3, -3)
2\pi	(5, -1)

The point starts at the rightmost point, reaches the top, then the left side, then the bottom: once counterclockwise around the circle with center (3, -1) and radius 2.

On 0 \le t \le \pi, \sin t \ge 0, so y \ge -1: only the upper half is traced, from (5, -1) over the top to (1, -1). With y = -1 - 2\sin t, the point at \displaystyle t = \frac{\pi}{2} is the bottom (3, -3), so the same circle is traced clockwise.

Example 5

Part 2: a toy train on a circular track

Adaugo’s toy train runs counterclockwise around a circular track of radius 2 meters, one lap every 24 seconds. Put the origin at the center and start the train at (2, 0). Write its position at time t seconds, find it at t = 8, and find its speed. Then do the same for a train that starts at the top, (0, 2), and runs clockwise at the same pace.

Solution

One lap is an angle of 2\pi in 24 seconds, so the angle after t seconds is \displaystyle \frac{2\pi t}{24} = \frac{\pi t}{12}:

x = 2\cos\frac{\pi t}{12}, \qquad y = 2\sin\frac{\pi t}{12}.

At t = 8 the angle is \displaystyle \frac{2\pi}{3}, so the train is at \displaystyle (-1, \sqrt{3}). In one lap it travels 2\pi \cdot 2 = 4\pi meters, so its speed is \displaystyle \frac{4\pi}{24} = \frac{\pi}{6} \approx 0.52 meters per second (compare Angular and linear speed).

For the second train, swap cosine and sine:

x = 2\sin\frac{\pi t}{12}, \qquad y = 2\cos\frac{\pi t}{12}.

At t = 0 this gives the top (0, 2), and at t = 6 the right side (2, 0): clockwise. The point stays 2 from the center and takes 24 seconds a lap. At t = 8 it is at \displaystyle (\sqrt{3}, -1).

Example 6

Part 2: one ellipse, three ways

Show that each pair traces the ellipse centered at the origin that reaches 5 units left and right and 3 units up and down, and say how: (a) x = 5\cos t, y = 3\sin t, 0 \le t \le 2\pi; (b) x = 5\cos 2t, y = 3\sin 2t, 0 \le t \le \pi; (c) x = 5\sin t, y = 3\cos t, 0 \le t \le 2\pi.

Solution

In every case \displaystyle \frac{x}{5} and \displaystyle \frac{y}{3} are a cosine and a sine of the same angle, so \displaystyle \left(\frac{x}{5}\right)^2 + \left(\frac{y}{3}\right)^2 = 1 and every point is on that ellipse.

(a) This is the ellipse form: (5, 0), (0, 3), (-5, 0), (0, -3) at the quarter turns, once counterclockwise from the right end.

(b) As t runs from 0 to \pi, 2t runs from 0 to 2\pi: the trip in (a) at double speed. The top comes at \displaystyle t = \frac{\pi}{4} instead of \displaystyle \frac{\pi}{2}.

(c) At t = 0 the point is the top (0, 3), and at \displaystyle t = \frac{\pi}{2} the right end (5, 0), so it goes around once clockwise, starting at the top.

A single equation in x and y cannot tell these three apart. Finding that equation from a parametric pair in general is Eliminating the parameter.

Example 7

Part 2: will the ball clear the fence?

Oluwatobi sets a pitching machine to launch a ball at 80 feet per second, 30^\circ above the horizontal, from 5 feet up. A fence 10 feet tall stands 170 feet away. Using the projectile model, find the highest point, decide whether the ball clears the fence, and find where it would land with no fence.

Solution

The velocity components are \displaystyle 80\cos 30^\circ = 40\sqrt{3} across and 80\sin 30^\circ = 40 up, so

x = 40\sqrt{3}\,t, \qquad y = 5 + 40t - 16t^2.

The height is a downward quadratic in t, largest at the vertex \displaystyle t = \frac{40}{2 \cdot 16} = 1.25 seconds, where y = 5 + 50 - 25 = 30 feet and \displaystyle x = 40\sqrt{3}(1.25) = 50\sqrt{3} \approx 86.6 feet.

The ball reaches the fence when x = 170, at \displaystyle t = \frac{170}{40\sqrt{3}} \approx 2.4537 seconds. Its height then is

y \approx 5 + 40(2.4537) - 16(2.4537)^2 \approx 6.8 \text{ feet},

less than 10. The ball hits the fence about 3.2 feet below the top.

Figure 3Example 7. The ball is still in the air at the fence, but too low to clear it. Past the fence, the path shows where it would land with no fence.

With no fence it lands when 16t^2 - 40t - 5 = 0. The quadratic formula gives \displaystyle t = \frac{5 \pm \sqrt{30}}{4}, and only the positive root comes after launch: t \approx 2.619 seconds, so \displaystyle x \approx 40\sqrt{3}(2.619) \approx 181.5 feet.

The parameter did real work: the x-equation found the moment the ball reached the fence, and the y-equation gave its height at that moment.

Boundaries

The interval is part of the curve. On a shorter interval the same equations trace only a piece (the half circle in Example 4); on a longer one they may retrace it (Exercise 13).
A curve does not determine its parametric equations (Examples 3 and 6). Orientation belongs to the equations, not to the picture.
The line form needs a and b not both zero; otherwise the point never moves.
\displaystyle \sqrt{a^2 + b^2} is the distance per unit of t only for the line form. On a curve the speed generally changes (the ball in Example 7 slows on the way up).
In the circle form r > 0 and t is in radians.
The projectile equations are a model. They ignore air resistance, so they overstate a real throw, and they stop applying once the ball lands. For launch and landing at the same height, the landing distance \displaystyle \frac{v_0^2 \sin 2\theta}{g} is found in Double angle formulas.

Where this goes next

The next lesson, Eliminating the parameter, goes from a parametric pair back to a single equation in x and y, which names the curve but drops the direction and timing. In calculus, Parametric, polar and vector valued functions finds the velocity of a moving point, slopes along a parametric curve, and the length of a path, with this lesson’s line and projectile as the first cases. The velocities of Forces and velocities as vectors become, with a parameter added, motion along a path.

Exercises

Part 1: tables and orientation

Make a table for x = 2t - 1, y = t^2 at t = -2, -1, 0, 1, 2. Give the initial and terminal points and the orientation.
Make a table for x = t^2, y = t^3 at t = -2, -1, 0, 1, 2. Describe the orientation.
Find the points of x = 4 - t, \displaystyle y = \sqrt{t} at t = 0, 1, 4, the initial and terminal points for 0 \le t \le 4, and the orientation.
The curve x = t^2 - 4, \displaystyle y = \frac{t}{2}, -3 \le t \le 3 crosses the y-axis twice. Find the values of t, the points, and the orientation.

Part 1: lines

For x = 2 + 3t, y = -2 + 4t, find the points at t = 0 and t = 2, and the distance moved per unit of t.
Write parametric equations for the segment from (-3, 5) to (5, -1) with 0 \le t \le 1. Find the point at \displaystyle t = \frac{1}{2} and the length of the segment.
Ngozika’s mower drives steadily in a straight line from (-6, 4) to (9, -4), in meters, in 10 seconds. Write its position at time t, its speed, and where and when it crosses the x-axis.
For x = 7 - 2t, y = -3 + 4t, decide whether each point is on the line: (a) (-1, 13); (b) (3, 2).
Write parametric equations that trace the segment of Exercise 6 from (5, -1) to (-3, 5) as t runs from 0 to 2. How far does the point move per unit of t?

Part 2: circles and ellipses

Describe the curve x = -2 + 3\cos t, y = 4 + 3\sin t, 0 \le t \le 2\pi: center, radius, the points at \displaystyle t = 0, \frac{\pi}{2}, \pi, and the orientation.
A circle has center (1, -3) and radius 5. (a) Parametrize it once counterclockwise from its rightmost point. (b) Parametrize it once counterclockwise from its top point.
Describe the curve x = 4\cos t, y = -4\sin t, 0 \le t \le 2\pi, including its orientation.
Describe the curve x = 6\cos 3t, y = 6\sin 3t, 0 \le t \le 2\pi. How many times is it traced, and how long does each lap take?
For x = 2 + 4\cos t, y = -1 + 2\sin t, 0 \le t \le 2\pi, find the points at \displaystyle t = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, the center, the total width and height, and the orientation.
Adaugo’s second track has radius 3 meters. A train starts at the bottom, (0, -3), and runs counterclockwise, one lap every 30 seconds. Write its position at time t, find it at t = 10, and find the speed.

Part 2: projectiles

Adaugo kicks a kickball from the ground at 14 meters per second, 30^\circ above the horizontal. Write the path, and find when and how far away it lands and its greatest height.
A ball is thrown from 6 feet up at 50 feet per second, 40^\circ above the horizontal. Write the path with decimal coefficients and find its position after 1 second.
A ball follows x = 24t, y = 4 + 32t - 16t^2 (feet, seconds). Find its launch height, speed and angle, and where it lands.
A stone is thrown horizontally at 12 meters per second from the top of a cliff 20 meters high. Write its path and find when and how far from the base of the cliff it lands.

Reasoning

Compare x = t, y = t^2; then x = t^3, y = t^6; then x = t^2, y = t^4, each for all real t. Which trace the whole parabola y = x^2, and with what orientation?
Show that every point of x = h + r\cos t, y = k + r\sin t is at distance r from (h, k), using the distance formula.
Explain why x = \cos t, y = \sin t goes counterclockwise while x = \sin t, y = \cos t goes clockwise.
In the projectile model, why does x have no t^2 term while y does? What would change with air resistance?
Explain why x = x_0 + at, y = y_0 + bt has slope \displaystyle \frac{b}{a} when a \ne 0, and describe the line when a = 0.

Answers

Points (-5, 4), (-3, 1), (-1, 0), (1, 1), (3, 4). Initial point (-5, 4), terminal point (3, 4). Since x = 2t - 1 increases with t, the parabola is traced left to right.
Points (4, -8), (1, -1), (0, 0), (1, 1), (4, 8). Since y = t^3 increases, the point always moves upward: in from the lower right to the origin at t = 0, then out to the upper right.
(4, 0), (3, 1), (0, 2). Initial point (4, 0), terminal point (0, 2). Since x = 4 - t decreases, the curve is traced right to left, rising.
x = 0 when t^2 = 4, so t = -2 or t = 2, giving (0, -1) and (0, 1). Since \displaystyle y = \frac{t}{2} increases, the curve is traced upward, from (5, -1.5) to (5, 1.5).
(2, -2) at t = 0 and (8, 6) at t = 2. Per unit of t it moves \displaystyle \sqrt{3^2 + 4^2} = 5.
x = -3 + 8t, y = 5 - 6t, 0 \le t \le 1. At \displaystyle t = \frac{1}{2}: (1, 2), the midpoint. Length \displaystyle \sqrt{8^2 + 6^2} = 10.
Rates \displaystyle \frac{15}{10} = 1.5 and \displaystyle \frac{-8}{10} = -0.8: x = -6 + 1.5t, y = 4 - 0.8t, 0 \le t \le 10. Speed \displaystyle \sqrt{1.5^2 + 0.8^2} = \sqrt{2.89} = 1.7 meters per second. y = 0 at t = 5, at the point (1.5, 0).
(a) x = -1 gives t = 4, and then y = -3 + 16 = 13: on the line. (b) x = 3 gives t = 2, and then y = 5 \ne 2: not on the line.
Change per unit of t: \displaystyle \frac{-8}{2} = -4 and \displaystyle \frac{6}{2} = 3, so x = 5 - 4t, y = -1 + 3t, 0 \le t \le 2. It moves \displaystyle \sqrt{16 + 9} = 5 per unit of t.
Center (-2, 4), radius 3. Points (1, 4), (-2, 7), (-5, 4): right side, top, left side, so once around counterclockwise from (1, 4).
(a) x = 1 + 5\cos t, y = -3 + 5\sin t, 0 \le t \le 2\pi. (b) x = 1 - 5\sin t, y = -3 + 5\cos t, 0 \le t \le 2\pi: at t = 0 it is at the top (1, 2), and at \displaystyle t = \frac{\pi}{2} at the left side (-4, -3), which is counterclockwise.
The circle of radius 4 centered at the origin, once, starting at (4, 0). At \displaystyle t = \frac{\pi}{2} it is at the bottom (0, -4), so it runs clockwise.
The circle of radius 6 centered at the origin. As t runs to 2\pi, 3t runs to 6\pi: three times counterclockwise, each lap taking \displaystyle \frac{2\pi}{3}.
(6, -1), (2, 1), (-2, -1), (2, -3). Center (2, -1), width 8, height 4. Counterclockwise from the right end.
The angle after t seconds is \displaystyle \frac{\pi t}{15}, measured from the bottom: \displaystyle x = 3\sin\frac{\pi t}{15}, \displaystyle y = -3\cos\frac{\pi t}{15} (at t = 0 this is (0, -3), at t = 7.5 it is (3, 0), so counterclockwise). At t = 10: \displaystyle \left(\frac{3\sqrt{3}}{2}, \frac{3}{2}\right). Speed \displaystyle \frac{6\pi}{30} = \frac{\pi}{5} \approx 0.63 meters per second.
\displaystyle 14\cos 30^\circ = 7\sqrt{3} and 14\sin 30^\circ = 7: \displaystyle x = 7\sqrt{3}\,t, y = 7t - 4.9t^2. Lands when t(7 - 4.9t) = 0, at \displaystyle t = \frac{10}{7} \approx 1.43 seconds, \displaystyle x = 10\sqrt{3} \approx 17.3 meters away. Highest at \displaystyle t = \frac{5}{7}: y = 5 - 2.5 = 2.5 meters.
50\cos 40^\circ \approx 38.30 and 50\sin 40^\circ \approx 32.14: x \approx 38.30t, y \approx 6 + 32.14t - 16t^2. At t = 1: about (38.30, 22.14).
Launch height 4 feet. Velocity \langle 24, 32 \rangle, speed \displaystyle \sqrt{24^2 + 32^2} = 40 feet per second, angle \displaystyle \arctan\frac{32}{24} \approx 53.13^\circ (an inverse tangent). Lands when 16t^2 - 32t - 4 = 0, at \displaystyle t = 1 + \frac{\sqrt{5}}{2} \approx 2.118 seconds, x \approx 50.83 feet out.
\theta = 0, so x = 12t, y = 20 - 4.9t^2. Lands when \displaystyle t^2 = \frac{20}{4.9}, \displaystyle t = \frac{10\sqrt{2}}{7} \approx 2.02 seconds, at \displaystyle x = \frac{120\sqrt{2}}{7} \approx 24.24 meters from the base.
In the first two, y = x^2 at every point and x takes every real value, so both trace the whole parabola, left to right, since t and t^3 both increase. In the third, x = t^2 \ge 0, so only the right half is traced: in from the far right to the origin as t rises to 0, then back out along the same half.
The distance from (h, k) is \displaystyle \sqrt{(r\cos t)^2 + (r\sin t)^2}, which equals \displaystyle \sqrt{r^2(\cos^2 t + \sin^2 t)}, which is \displaystyle \sqrt{r^2} = r since r > 0.
The first starts at (1, 0) and reaches (0, 1) at \displaystyle t = \frac{\pi}{2}: right side to top is a counterclockwise quarter turn. The second starts at (0, 1) and reaches (1, 0): top to right side is clockwise. Swapping x and y reflects across y = x, which reverses the direction of turning.
A t^2 term comes from a steady pull changing the speed. Gravity pulls only straight down, so only the vertical speed changes; with no air resistance the horizontal speed stays v_0\cos\theta. Air resistance pushes against the motion, horizontally too, so x would grow more and more slowly and the simple formula for x would fail.
Between times t_1 and t_2 the rise is b(t_2 - t_1) and the run is a(t_2 - t_1), so the slope is \displaystyle \frac{b}{a} for every pair of points: a line. If a = 0 (and b \ne 0), x = x_0 never changes, so the curve is the vertical line x = x_0, which has no slope.