AP Precalculus: Parametric Functions
Each conic has a standard parametrization. Parabola. Solve for whichever variable is easiest and let the other be t. From y = ax² use (t, at²); from x = ay² use (ay², y) = (at², t). For example, y² = 12x becomes x = t²/12, y = t (or x = 3t², y = 6t). Ellipse. For (x − h)²/a² + (y − k)²/b² = 1 use x = h + a cos t, y = k + b sin t, 0 ≤ t ≤ 2π. The number under the x-term, a², gives the multiplier on the cosine, and b² goes with the sine, regardless of which one is larger. This works because ((a cos t)/a)² + ((b sin t)/b)² = cos² t + sin² t = 1. Hyperbola. Because sec² t − tan² t = 1, a horizontal hyperbola (x − h)²/a² − (y − k)²/b² = 1 is x = h + a sec t, y = k + b tan t, and a vertical hyperbola (y − k)²/a² − (x − h)²/b² = 1 is x = h + b tan t, y = k + a sec t. These formulas are undefined where cos t = 0 (at t = π/2 and 3π/2), which is exactly the gap between the two branches. To check any parametrization, substitute x(t) and y(t) back into the equation and simplify to a true statement. Remember too that these answers are not unique: replacing t with −t reverses the direction, and replacing t with 2t doubles the speed.
1. Parametric Equations: Tables, Graphs, and Direction
Sometimes a single equation in x and y is not enough. Picture a drone flying along the path y = x² − 3x. The equation tells you where the drone has been, but not when it was at each place. To capture time, add a third variable called the parameter, usually t, and write both coordinates as functions of it: x = x(t) and y = y(t). These are the parametric equations, and the pair together can be written as the single function f(t) = (x(t), y(t)). Each value of t in the stated domain gives one point (x(t), y(t)), and the set of all those points is the plane curve. To graph it by hand, build a table with columns t, x, y, plot the points, and connect them in order of increasing t. The order matters because a parametric curve has a direction of motion, so draw arrows along it. If the problem gives an interval such as −1 ≤ t ≤ 4, the curve has a starting point and an ending point at the endpoints. If no interval is given, assume t can be any real number and the curve continues in both directions. Graphing calculators have a parametric mode (MODE, then PAR) where you type x(T) and y(T), set Tmin, Tmax, and a small Tstep, and watch the curve trace.
Worked Example: Make a table and sketch the curve x(t) = t² − 2t, y(t) = t + 1 for −1 ≤ t ≤ 4. Then describe the motion.
Substitute each integer t: t = −1 gives (3, 0); t = 0 gives (0, 1); t = 1 gives (−1, 2); t = 2 gives (0, 3); t = 3 gives (3, 4); t = 4 gives (8, 5). Plot these and connect them in that order.
Because y = t + 1 rises steadily, the particle always moves up. Because x = (t − 1)² − 1 has its lowest value −1 at t = 1, the particle moves left for −1 < t < 1 and right for 1 < t < 4. The path is a sideways parabola whose leftmost point is (−1, 2).
start (3, 0), end (8, 5), vertex (−1, 2)
1. A particle’s position is x(t) = 2t + 1, y(t) = t² − 3. Where is it at t = 2?
Explanation: x(2) = 2(2) + 1 = 5 and y(2) = 2² − 3 = 1, so the position is (5, 1). Be careful to list x first and y second.
2. For x(t) = 4 − t and y(t) = t, in which direction does the particle move as t increases?
Explanation: As t increases, x = 4 − t decreases (left) and y = t increases (up). The path is the line x + y = 4, traced from lower right to upper left.
3. The curve x(t) = t + 2, y(t) = t² for −1 ≤ t ≤ 2 has which starting and ending points?
Explanation: At t = −1: x = 1, y = 1. At t = 2: x = 4, y = 4. The endpoints come from substituting the endpoint values of t into both equations.
4. On the curve x(t) = 3t + 1, y(t) = t² + 7, which value of t gives the point (7, 11)?
Explanation: Solve 3t + 1 = 7 to get t = 2, then check the y-value: 2² + 7 = 11. Both coordinates must match at the same t.
2. Parametric Functions Modeling Planar Motion
A parametric function f(t) = (x(t), y(t)) is the natural model for a particle moving in the plane: t is time and (x(t), y(t)) is its position. Questions about the motion can be answered by looking at x(t) and y(t) separately. Extrema: the horizontal extremes of the path (how far left and right it goes) are the maximum and minimum values of x(t), and the vertical extremes (how far down and up) are the maximum and minimum of y(t), taken over the given domain. Remember to check the endpoints of the interval as well as the vertex of any quadratic. Zeros: the real zeros of x(t) are the times when the particle is on the y-axis, so they produce the y-intercepts; the real zeros of y(t) are the times when the particle is on the x-axis, so they produce the x-intercepts. To find an intercept, solve the right equation for t, then substitute that t into the other function. A particle can cross an axis more than once, and only values of t inside the domain count. Position at a given time is just a substitution, and values far outside a table can be found directly, for example x(10) and y(10).
Worked Example: A particle moves according to x(t) = t² − 6t + 5, y(t) = 3 − t for 0 ≤ t ≤ 5. Find its horizontal and vertical extremes and its intercepts.
Horizontal extremes. x(t) = (t − 3)² − 4 has its vertex at t = 3, so the minimum is −4. Compare the endpoints: x(0) = 5 and x(5) = 0. So x ranges from −4 to 5.
Vertical extremes. y(t) = 3 − t is decreasing from y(0) = 3 to y(5) = −2, so y ranges from −2 to 3.
y-intercepts (x = 0). t² − 6t + 5 = (t − 1)(t − 5) = 0 gives t = 1 and t = 5. Then y(1) = 2 and y(5) = −2, so the intercepts are (0, 2) and (0, −2).
x-intercept (y = 0). 3 − t = 0 gives t = 3, and x(3) = −4. So the intercept is (−4, 0).
1. For x(t) = t² − 4t on 0 ≤ t ≤ 5, what are the greatest and least values of x?
Explanation: x(t) = (t − 2)² − 4, so the minimum is −4 at t = 2. The endpoints give x(0) = 0 and x(5) = 5, so the maximum is 5. Never ignore the endpoints.
2. What are the x-intercepts of x(t) = 3t − 6, y(t) = t² − 4?
Explanation: x-intercepts occur when y = 0: t² − 4 = 0, so t = 2 or t = −2. Then x(2) = 0 and x(−2) = −12. The intercepts are (0, 0) and (−12, 0), not the t-values.
3. What are the y-intercepts of x(t) = t² − 9, y(t) = 2t + 1?
Explanation: y-intercepts occur when x = 0: t² = 9, so t = 3 or t = −3. Then y(3) = 7 and y(−3) = −5.
4. For x(t) = 3t and y(t) = 4 − (t − 1)² on 0 ≤ t ≤ 3, what are the vertical maximum and minimum?
Explanation: y has its vertex (a maximum) at t = 1, where y = 4. The endpoints give y(0) = 3 and y(3) = 4 − 4 = 0, so the minimum is 0 at t = 3.
3. Direction of Motion and Rates of Change
The direction a particle moves can be read from x(t) and y(t) one at a time. If x(t) is increasing, the particle moves to the right; if x(t) is decreasing, it moves to the left. If y(t) is increasing, it moves up; if y(t) is decreasing, it moves down. Combining the two gives the diagonal directions, and the direction can change from one interval of t to another, as you will see on a circle or ellipse, where it turns continuously. For the amount of change, use the average rate of change over an interval [t1, t2]: for x it is Δx/Δt = (x(t2) − x(t1))/(t2 − t1), and for y it is Δy/Δt, and each has units such as meters per second. The two rates together tell you the slope of the secant line between the two points on the curve: slope = (Δy/Δt)÷(Δx/Δt) = Δy/Δx, provided Δx/Δt ≠ 0. If Δx/Δt = 0, the two points are directly above one another and the slope is undefined (a vertical segment). If Δy/Δt = 0 but Δx/Δt ≠ 0, the segment is horizontal with slope 0.
Worked Example (direction): Describe the motion of x(t) = 4 cos t, y(t) = 3 sin t for 0 ≤ t ≤ 2π.
On (0, π/2), cosine is decreasing and sine is increasing, so the particle moves left and up. On (π/2, π), both are decreasing: left and down. On (π, 3π/2), cosine increases while sine decreases: right and down. On (3π/2, 2π), both increase: right and up. The particle starts at (4, 0) and travels once around the ellipse counterclockwise.
Worked Example (rates): For x(t) = t², y(t) = t³ − 3t on [0, 2], find the average rates of change and the slope of the secant line.
x(0) = 0 and x(2) = 4, so Δx/Δt = 4/2 = 2. y(0) = 0 and y(2) = 8 − 6 = 2, so Δy/Δt = 2/2 = 1. The slope between the points (0, 0) and (4, 2) is 1/2, which equals (1)÷(2).
1. For x(t) = cos t and y(t) = sin t with π/2 < t < π, which direction does the particle move?
Explanation: On (π/2, π) both cos t and sin t are decreasing (cosine falls from 0 to −1, sine falls from 1 to 0), so the particle moves left and down.
2. Find the average rate of change of x(t) = t² + 1 on [1, 4].
Explanation: x(4) = 17 and x(1) = 2, so (17 − 2)/(4 − 1) = 15/3 = 5. The number 15 is only the change in x, not the rate.
3. For x(t) = 2t and y(t) = t² − 1 on [0, 3], what is the slope of the secant line between the endpoints?
Explanation: x(0) = 0, x(3) = 6, y(0) = −1, and y(3) = 8. So Δx = 6 and Δy = 9, and the slope is 9/6 = 3/2. Equivalently, the average rates are Δx/Δt = 2 and Δy/Δt = 3, and 3÷2 = 3/2.
4. Over an interval, the average rate of change of x(t) is 0 and that of y(t) is 5. What is true of the secant line?
Explanation: Both points have the same x-value, so they lie directly above one another. The slope Δy/Δx would require dividing by 0, so it is undefined.
4. Parametrizing Lines and Circles
A line segment. A straight path can be parametrized by saying where the particle is at t = 0 and t = 1. Let the start be (x0, y0) and the position one unit of time later be (x1, y1). Compute the rates Δx/Δt = x1 − x0 and Δy/Δt = y1 − y0. Then x = x0 + (Δx/Δt)t and y = y0 + (Δy/Δt)t. This is just the point-slope idea with constant velocity: you start at (x0, y0) and add a fixed change each unit of time. If t is allowed to take all real values, the graph is the entire line; if 0 ≤ t ≤ 1, it is exactly the segment between the two points. A circle. The unit circle x² + y² = 1 follows from cos² t + sin² t = 1, so x = cos t, y = sin t for 0 ≤ t ≤ 2π traces it once, counterclockwise from (1, 0). A circle of radius a centered at the origin is x = a cos t, y = a sin t, and a circle with center (h, k) is x = h + a cos t, y = k + a sin t: the h and k shift the position while the cosine and sine create the turning. Changing the sign of the sine term (y = k − a sin t) makes the circle clockwise, and replacing t with 2t makes the particle go around twice as fast.
Worked Example: (a) Write parametric equations for the segment from A(−3, 5) to B(1, −1). (b) Parametrize the circle (x + 2)² + (y − 1)² = 25.
(a) Start at A when t = 0 and reach B when t = 1. Δx/Δt = 1 − (−3) = 4 and Δy/Δt = −1 − 5 = −6. So x = −3 + 4t and y = 5 − 6t for 0 ≤ t ≤ 1. Check t = 1: (1, −1). Correct.
(b) The center is (−2, 1) and the radius is √25 = 5, so x = −2 + 5 cos t and y = 1 + 5 sin t for 0 ≤ t ≤ 2π. Check t = 0: (3, 1), and (3 + 2)² + 0 = 25.
1. Which parametrization gives the segment from (2, 7) at t = 0 to (6, 3) at t = 1?
Explanation: Δx/Δt = 6 − 2 = 4 and Δy/Δt = 3 − 7 = −4. Start at (2, 7) and add those amounts: x = 2 + 4t, y = 7 − 4t.
2. Which parametrization gives the circle (x − 3)² + (y + 1)² = 16?
Explanation: The center is (3, −1) and the radius is 4. The center values come from h = 3 and k = −1 (note the signs), and the radius 4, not 16, multiplies the cosine and sine.
3. The curve x = −4 + 7 cos t, y = 2 + 7 sin t, 0 ≤ t ≤ 2π, is a circle with
Explanation: x = h + a cos t and y = k + a sin t, so h = −4, k = 2 and a = 7.
4. The segment from (1, 1) at t = 0 to (4, 5) at t = 1 is extended to all real t. Where is the particle at t = 2?
Explanation: x = 1 + 3t and y = 1 + 4t. At t = 2: x = 7 and y = 9. The particle is on the same line, beyond the segment.
5. Implicitly Defined Functions
An implicitly defined equation relates x and y without solving for one in terms of the other. The circle x² + y² = 25 is an example. Many implicit equations can be solved for one variable, but not all: an equation like y + arctan y = x cannot be written as y = (something in x) with ordinary algebra. To graph an implicit equation, substitute a chosen value for one variable and solve for the other, then plot the ordered pairs. The drawback is that you may need many points, and short stretches (such as sudden steepness near an intercept) can be missed. Technology helps: solve for y when you can and graph the explicit pieces. Solving for y in x² + y² = 25 gives y = ±√(25 − x²), two functions whose graphs are the upper and lower semicircles. Solving for x gives x = ±√(25 − y²), the right and left semicircles. Solving for one variable defines a function whose graph is part of the implicit graph. Rate of change. For points that are close on the curve, if the average rate of change Δy/Δx is positive, the two variables increase or decrease together; if it is negative, one increases while the other decreases; if it is zero, the stretch is horizontal; and if x does not change while y does, the stretch is vertical (the rate is undefined).
Worked Example: For x² + y² = 25, (a) write y explicitly, (b) find the average rate of change of y with respect to x for 0 ≤ x ≤ 3 with y ≥ 0, and (c) do the same for −5 ≤ x ≤ −3 with y ≥ 0.
(a) y² = 25 − x², so y = ±√(25 − x²).
(b) With y ≥ 0, x = 0 gives y = 5 and x = 3 gives y = 4 (reject −4). So Δy/Δx = (4 − 5)/(3 − 0) = −1/3. The rate is negative: as x increases, y decreases.
(c) x = −5 gives y = 0 and x = −3 gives y = 4. So Δy/Δx = (4 − 0)/(−3 − (−5)) = 4/2 = 2. The rate is positive: y increases as x increases on the upper left part of the circle.
1. Solving x² + y² = 36 for y gives
Explanation: Subtract x² from both sides: y² = 36 − x², then take both square roots. You cannot take the square root of each term separately.
2. Which equation graphs the right half of x² + y² = 49?
Explanation: Solving for x gives x = ±√(49 − y²). The positive root has x ≥ 0, so it is the right half.
3. On the part of x² + y² = 100 with y ≥ 0, what is the average rate of change of y with respect to x on [0, 6]?
Explanation: y(0) = 10 and y(6) = √(100 − 36) = 8. The rate is (8 − 10)/6 = −1/3.
4. On the quarter of x² + y² = 100 in Quadrant II, what is the average rate of change of y with respect to x on [−10, 0]?
Explanation: y(−10) = 0 and y(0) = 10, so (10 − 0)/(0 − (−10)) = 1. It is positive because y increases as x increases in Quadrant II.
6. Conic Sections: The Parabola
Circles, ellipses, parabolas, and hyperbolas are called conic sections because each is a cross-section of a double cone. A plane perpendicular to the axis cuts a circle; a tilted plane that cuts only one cone makes an ellipse; a plane parallel to the edge of the cone makes a parabola; and a plane that cuts both cones makes a hyperbola. A parabola is defined geometrically as the set of all points the same distance from a fixed point (the focus) and a fixed line (the directrix). The line through the focus perpendicular to the directrix is the axis of symmetry (focal axis), and the vertex is where the parabola crosses it, halfway between the focus and the directrix. Let p be the signed distance from the vertex to the focus. With vertex (h, k) there are two standard forms. Vertical axis: (x − h)² = 4p(y − k), opening up if p > 0 and down if p < 0, with focus (h, k + p), directrix y = k − p, and axis x = h. Horizontal axis: (y − k)² = 4p(x − h), opening right if p > 0 and left if p < 0, with focus (h + p, k), directrix x = h − p, and axis y = k. The squared variable tells you the orientation: if x is squared, the parabola opens up or down; if y is squared, it opens left or right. Writing the equation as a(y − k) = (x − h)² makes a = 4p.
Worked Example: (a) Find the vertex, direction, focus and directrix of (x + 3)² = 12(y − 2). (b) Do the same for (y − 1)² = −8(x + 2). (c) Write the equation of the parabola with vertex (1, −2) that opens right and passes through (3, 2).
(a) The vertex is (−3, 2). Since x is squared and 4p = 12, p = 3 > 0, so it opens up. The focus is (−3, 2 + 3) = (−3, 5), the directrix is y = 2 − 3 = −1, and the axis is x = −3.
(b) The vertex is (−2, 1). Since y is squared and 4p = −8, p = −2 < 0, so it opens left. The focus is (−2 + (−2), 1) = (−4, 1), the directrix is x = −2 − (−2) = 0, and the axis is y = 1.
(c) A parabola opening right with vertex (1, −2) has the form (y + 2)² = a(x − 1). Substitute (3, 2): (2 + 2)² = a(3 − 1), so 16 = 2a and a = 8. The equation is (y + 2)² = 8(x − 1), with p = 2 and focus (3, −2).
1. What are the vertex and direction of (x − 4)² = −12(y + 1)?
Explanation: Match (x − h)² = 4p(y − k): h = 4 and k = −1 (note y + 1 = y − (−1)). 4p = −12 is negative and x is squared, so it opens down.
2. What is the focus of x² = 16y?
Explanation: The vertex is (0, 0) and 4p = 16, so p = 4. The parabola opens up, so the focus is p units above the vertex at (0, 4).
3. What is the focus of (y − 3)² = 20(x + 1)?
Explanation: The vertex is (−1, 3) and 4p = 20, so p = 5. The axis is horizontal (y is squared) and it opens right, so the focus is (−1 + 5, 3) = (4, 3).
4. What is the directrix of (y − 3)² = 20(x + 1)?
Explanation: The directrix is on the opposite side of the vertex from the focus, p = 5 units away: x = −1 − 5 = −6.
5. Which equation describes the parabola with vertex (1, 2) that opens up and passes through (3, 3)?
Explanation: Use (x − 1)² = a(y − 2) and substitute (3, 3): 4 = a(1), so a = 4. Choice D has the squared variable on the wrong side, making the parabola open right.
7. Conic Sections: The Ellipse
An ellipse is the set of all points in a plane for which the sum of the distances to two fixed points, the foci, is constant. Think of two thumbtacks and a loop of string. The line through the foci is the focal axis. The midpoint of the foci is the center. The two points where the ellipse crosses the focal axis are the vertices, and the segment joining them is the major axis of length 2a. The perpendicular segment through the center that ends on the ellipse is the minor axis of length 2b, with a > b. The three lengths are linked by a² = b² + c², so c = √(a² − b²) is the distance from the center to each focus. The standard equation has 1 alone on the right side. Horizontal major axis: (x − h)²/a² + (y − k)²/b² = 1, with foci (h ± c, k) and vertices (h ± a, k). Vertical major axis: (x − h)²/b² + (y − k)²/a² = 1, with foci (h, k ± c) and vertices (h, k ± a). To decide the orientation, look at which denominator is larger: the larger one belongs to the variable along the major axis, and a² is always the larger denominator. If a = b the ellipse is a circle (c = 0). An equation like 4x² + 9y² = 36 must first be divided by 36 to reach standard form.
Worked Example: (a) Describe x²/25 + y²/9 = 1. (b) Describe (x − 2)²/16 + (y + 1)²/25 = 1. (c) An ellipse has foci (−3, 4) and (5, 4) and a major axis of length 12. Write its equation.
(a) 25 > 9 and 25 is under x, so the major axis is horizontal. a = 5, b = 3, c = √(25 − 9) = 4. Center (0, 0), vertices (±5, 0), foci (±4, 0), minor-axis endpoints (0, ±3). Major axis 10, minor axis 6.
(b) 25 > 16 and 25 is under y, so the major axis is vertical. a = 5, b = 4, c = 3. Center (2, −1), vertices (2, −1 ± 5) = (2, 4) and (2, −6), foci (2, −1 ± 3) = (2, 2) and (2, −4).
(c) The foci have the same y-coordinate, so the major axis is horizontal. The center is the midpoint: ((−3 + 5)/2, 4) = (1, 4), and c = 4. The major axis is 12, so a = 6, and b² = a² − c² = 36 − 16 = 20. The equation is (x − 1)²/36 + (y − 4)²/20 = 1.
1. What are the foci of x²/49 + y²/25 = 1?
Explanation: The larger denominator, 49, is under x, so the major axis is horizontal and c = √(49 − 25) = √24 = 2√6. The foci are (±2√6, 0). The value √74 comes from adding instead of subtracting.
2. What is the length of the major axis of x²/16 + y²/36 = 1?
Explanation: The larger denominator is 36, so a = 6 and the major axis (vertical) has length 2a = 12. The minor axis has length 8.
3. An ellipse has foci (0, ±5) and vertices (0, ±13). What is its equation?
Explanation: The major axis is vertical with a = 13 and c = 5, so b² = 169 − 25 = 144. The larger denominator, 169, goes under y.
4. What are the vertices of 9x² + 4y² = 36?
Explanation: Divide by 36: x²/4 + y²/9 = 1. The larger denominator 9 is under y, so the major axis is vertical with a = 3, and the vertices are (0, ±3).
5. What is the center of (x + 5)²/9 + (y − 2)²/4 = 1?
Explanation: x + 5 = x − (−5), so h = −5 and k = 2.
8. Conic Sections: The Hyperbola
A hyperbola is the set of all points for which the difference of the distances to two fixed foci is constant. It has two separate branches that mirror each other. The center is the midpoint of the foci, the vertices are the points on each branch closest to the center (a units away), and the foci are c units from the center, so c > a. The segment between the vertices is the transverse axis (length 2a) and the perpendicular segment of length 2b is the conjugate axis. The three lengths satisfy c² = a² + b² (note the plus sign, unlike the ellipse). The asymptotes are the lines through the center that contain the diagonals of the rectangle with sides 2a and 2b; the branches get closer and closer to them. Horizontal hyperbola: (x − h)²/a² − (y − k)²/b² = 1, vertices (h ± a, k), foci (h ± c, k), asymptotes y − k = ±(b/a)(x − h). Vertical hyperbola: (y − k)²/a² − (x − h)²/b² = 1, vertices (h, k ± a), foci (h, k ± c), asymptotes y − k = ±(a/b)(x − h). The positive term tells the orientation: the variable whose term is positive (comes first) lies along the transverse axis, and a² is the denominator of that positive term. Unlike the ellipse, a can be larger than, smaller than, or equal to b. Always rewrite the equation in standard form (1 on the right) first.
Worked Example: (a) List the key features of 16x² − 9y² = 144. (b) Describe (y − 1)²/9 − (x + 2)²/16 = 1. (c) Write the equation of the hyperbola with vertices (±6, 0) and foci (±10, 0).
(a) Divide by 144: x²/9 − y²/16 = 1. The x-term is positive, so it is horizontal with center (0, 0). a = 3 and b = 4, so c = √(9 + 16) = 5. Vertices (±3, 0), foci (±5, 0), transverse axis 6, conjugate axis 8, asymptotes y = ±(4/3)x.
(b) The y-term is positive, so it is vertical with center (−2, 1). a = 3, b = 4, c = 5. Vertices (−2, 1 ± 3) = (−2, 4) and (−2, −2). Foci (−2, 1 ± 5) = (−2, 6) and (−2, −4). Asymptotes y − 1 = ±(3/4)(x + 2).
(c) Horizontal, center (0, 0), a = 6, c = 10, so b² = c² − a² = 100 − 36 = 64. The equation is x²/36 − y²/64 = 1.
1. What are the asymptotes of 25x² − 9y² = 225?
Explanation: Divide by 225: x²/9 − y²/25 = 1, so a = 3 and b = 5. For a horizontal hyperbola the asymptotes are y = ±(b/a)x = ±(5/3)x.
2. What are the foci of x²/20 − y²/5 = 1?
Explanation: For a hyperbola c² = a² + b² = 20 + 5 = 25, so c = 5. The hyperbola is horizontal, so the foci are (±5, 0). √15 comes from subtracting, which is for ellipses.
3. What are the vertices of y²/16 − x²/9 = 1?
Explanation: The y-term is positive, so the transverse axis is vertical and a² = 16, a = 4. The vertices are (0, ±4). The foci would be at (0, ±5).
4. A hyperbola centered at the origin has vertices (±4, 0) and foci (±5, 0). What is its equation?
Explanation: a = 4 and c = 5, so b² = 25 − 16 = 9. The vertices are on the x-axis, so the x-term is positive: x²/16 − y²/9 = 1.
5. What type of conic is 4y² − x² = 36?
Explanation: Divide by 36: y²/9 − x²/36 = 1. A minus sign between the squared terms means a hyperbola, and since y² is positive, it opens up and down.
9. Parametrizing Functions and Their Inverses
Any function y = f(x) can be parametrized by letting x = t: (x(t), y(t)) = (t, f(t)), with the domain of t the same as the domain of f. If there is no restriction, t can be any real number. A parametrization is not unique: letting x = 3t − 2 (as long as it covers all x-values in the domain) and substituting into f gives a different pair of equations that traces the same curve, and the same curve can even be traced in the opposite direction or at a different speed. The check is always that substituting x(t) and y(t) into the original equation gives a true statement for every t. Inverses. A function is one-to-one if different inputs always give different outputs. On a graph, the horizontal line test detects this: if every horizontal line meets the graph at most once, f is one-to-one and has an inverse that is also a function. If f is invertible, the curve of f−1 is found by swapping the roles of x and y: parametrize f as (t, f(t)) and its inverse as (f(t), t), or solve for the inverse function algebraically (replace f(x) by y, swap x and y, solve for y) and parametrize that as (t, f−1(t)). The graph of the inverse is the reflection of f over the line y = x. If a function is not one-to-one, restrict its domain to an interval where it is, such as x ≥ 2 for a parabola with vertex at x = 2, and then find the inverse of the restricted function.
Worked Example: (a) Give two parametrizations of y = x² + 4x. (b) For f(x) = x³ + 5, find f−1 and parametrize both. (c) Show that g(x) = x² − 4x is not one-to-one, then restrict it to find an inverse.
(a) First: x = t, y = t² + 4t. Second: x = 2t + 1, which covers every real x, so y = (2t + 1)² + 4(2t + 1) = 4t² + 12t + 5. Check t = 0: (1, 5), and 1² + 4(1) = 5.
(b) Swap and solve: x = y³ + 5, so y = ∛(x − 5), giving f−1(x) = ∛(x − 5). Parametrize f as (t, t³ + 5). The inverse can be (t³ + 5, t) or (t, ∛(t − 5)); both trace the same curve.
(c) g(1) = −3 and g(3) = −3, so two inputs give the same output and g is not one-to-one. The vertex is at x = 2, so restrict to x ≥ 2. Write g(x) = (x − 2)² − 4, swap: x = (y − 2)² − 4, so (y − 2)² = x + 4, and since y ≥ 2 we choose the positive root: g−1(x) = 2 + √(x + 4) for x ≥ −4.
1. Which gives a valid parametrization of y = 3x² − 2?
Explanation: Set x = t and substitute t for x in the formula: y = 3t² − 2. Each other choice fails when you substitute back, for example x = 3t, y = t² − 2 gives y = x²/9 − 2.
2. If y = x² is parametrized with x(t) = t + 2, what is y(t)?
Explanation: Replace x with the expression t + 2 in y = x²: y = (t + 2)² = t² + 4t + 4. Option t² + 2 is a common error that squares only the t.
3. Which function is one-to-one?
Explanation: x³ − 1 is always increasing, so it passes the horizontal line test. The others give the same output for x and −x.
4. Which parametrizes the inverse of f(x) = 2x − 5?
Explanation: The inverse of (t, f(t)) is (f(t), t) = (2t − 5, t). Algebraically, the inverse function is y = (x + 5)/2, and (2t − 5, t) satisfies it: (2t − 5 + 5)/2 = t.
5. Which domain restriction makes g(x) = x² − 4x one-to-one?
Explanation: The vertex of g is at x = 2, so the function is one-to-one on x ≥ 2 (or x ≤ 2). The other intervals all contain both 1 and 3, which give the same output g(1) = g(3) = −3.
10. Eliminating the Parameter: Parametric to Rectangular Form
To convert parametric equations to a rectangular (x, y) equation, eliminate the parameter. There are two standard techniques. Algebraic substitution: solve one equation for t, then substitute into the other. This works whenever one of the equations is easy to solve for t, such as x = 5 − 2t. Trigonometric identities: when x and y involve sine and cosine, isolate cos t and sin t in each equation and use cos² t + sin² t = 1 (or, with secant and tangent, sec² t − tan² t = 1). The result is usually a conic, so you can recognize the curve right away: x = a cos t and y = b sin t give x²/a² + y²/b² = 1, an ellipse. Do not forget the domain. The original interval of t may limit the part of the rectangular graph that is actually drawn. Find the x-values and y-values the parametrization actually reaches. For example, a parabola segment may have endpoints, and x = cos t, y = sin t on [0, π] is only the upper half of the unit circle because sin t ≥ 0 there. The rectangular equation x² + y² = 1 by itself would wrongly show the whole circle. Also, the rectangular equation does not show direction, so use the table or arrows from the parametric form.
Worked Example: Eliminate the parameter. (a) x = 5 − 2t, y = t² + 3, −1 ≤ t ≤ 4. (b) x = 4 sin t, y = 2 cos t, 0 ≤ t ≤ 2π. (c) x = 3 cos t, y = 3 sin t, 0 ≤ t ≤ π. (d) x = t² + 1, y = t, t ≥ 0.
(a) Solve for t: t = (5 − x)/2. Then y = ((5 − x)/2)² + 3 = (x² − 10x + 25)/4 + 3 = (x² − 10x + 37)/4. Domain: t from −1 to 4 gives x from 7 down to −3, so −3 ≤ x ≤ 7. Range: y = t² + 3 gives 3 ≤ y ≤ 19.
(b) sin t = x/4 and cos t = y/2. Then (x/4)² + (y/2)² = 1, or x²/16 + y²/4 = 1: an ellipse centered at the origin traced once clockwise (it starts at (0, 2) when t = 0 and goes right).
(c) cos t = x/3 and sin t = y/3 give x² + y² = 9. Because 0 ≤ t ≤ π makes sin t ≥ 0, only the upper half, y ≥ 0, is traced.
(d) y = t, so x = y² + 1. Because t ≥ 0, we have y ≥ 0: the upper half of a sideways parabola with vertex (1, 0).
1. Eliminate the parameter: x = t + 3, y = t².
Explanation: Solve t = x − 3, then substitute: y = (x − 3)². The option y = x² − 3 only subtracts 3 outside the square.
2. Eliminate the parameter: x = 2 cos t, y = 5 sin t, 0 ≤ t ≤ 2π.
Explanation: cos t = x/2 and sin t = y/5. Squaring and adding: x²/4 + y²/25 = 1, an ellipse with a vertical major axis.
3. What curve is x = 6 cos t, y = 6 sin t for 0 ≤ t ≤ π?
Explanation: x² + y² = 36 is the circle, but on [0, π] sin t ≥ 0, so y ≥ 0. Only the upper semicircle is traced.
4. Find the line for x = 2t + 1, y = 3t − 4.
Explanation: t = (x − 1)/2, so y = 3(x − 1)/2 − 4 = (3/2)x − 3/2 − 4 = (3/2)x − 11/2.
5. Which describes x = t² + 1, y = t for t ≥ 0?
Explanation: Since y = t, replace t with y: x = y² + 1. The condition t ≥ 0 becomes y ≥ 0, so only the upper half of the parabola is traced.
11. Representing Conic Sections Parametrically
Each conic has a standard parametrization. Parabola. Solve for whichever variable is easiest and let the other be t. From y = ax² use (t, at²); from x = ay² use (ay², y) = (at², t). For example, y² = 12x becomes x = t²/12, y = t (or x = 3t², y = 6t). Ellipse. For (x − h)²/a² + (y − k)²/b² = 1 use x = h + a cos t, y = k + b sin t, 0 ≤ t ≤ 2π. The number under the x-term, a², gives the multiplier on the cosine, and b² goes with the sine, regardless of which one is larger. This works because ((a cos t)/a)² + ((b sin t)/b)² = cos² t + sin² t = 1. Hyperbola. Because sec² t − tan² t = 1, a horizontal hyperbola (x − h)²/a² − (y − k)²/b² = 1 is x = h + a sec t, y = k + b tan t, and a vertical hyperbola (y − k)²/a² − (x − h)²/b² = 1 is x = h + b tan t, y = k + a sec t. These formulas are undefined where cos t = 0 (at t = π/2 and 3π/2), which is exactly the gap between the two branches. To check any parametrization, substitute x(t) and y(t) back into the equation and simplify to a true statement. Remember too that these answers are not unique: replacing t with −t reverses the direction, and replacing t with 2t doubles the speed.
Worked Example: Parametrize (a) the parabola x² = −12y, (b) the ellipse (x − 3)²/49 + (y + 2)²/16 = 1, (c) the hyperbola (x + 1)²/25 − (y − 4)²/9 = 1, (d) the hyperbola y²/4 − x²/36 = 1.
(a) Solve for y: y = −x²/12. Let x = t: (x(t), y(t)) = (t, −t²/12).
(b) h = 3, k = −2, a = 7, b = 4: x = 3 + 7 cos t, y = −2 + 4 sin t.
(c) Horizontal: h = −1, k = 4, a = 5, b = 3: x = −1 + 5 sec t, y = 4 + 3 tan t. Check: (5 sec t)²/25 − (3 tan t)²/9 = sec² t − tan² t = 1.
(d) Vertical: a² = 4 and b² = 36 so a = 2, b = 6, center (0, 0): x = 6 tan t, y = 2 sec t.
1. Which parametrization of the parabola y² = 8x is correct?
Explanation: Solve for x: x = y²/8, and let y = t, so x = t²/8. Check: y² = t² and 8x = t². The other choices do not satisfy y² = 8x for all t.
2. Which parametrizes (x + 2)²/9 + (y − 5)²/25 = 1?
Explanation: The center is (−2, 5), a = √9 = 3 under the x-term, and b = √25 = 5 under the y-term. The multipliers are the square roots of the denominators, not the denominators themselves.
3. Which parametrizes the hyperbola (x − 1)²/16 − (y + 2)²/4 = 1?
Explanation: The x-term is positive (horizontal), a = 4, b = 2, h = 1, k = −2. So x = h + a sec t and y = k + b tan t. Check: sec² t − tan² t = 1.
4. For x = 4 + 6 cos t, y = −1 + 3 sin t, which is the rectangular equation?
Explanation: cos t = (x − 4)/6 and sin t = (y + 1)/3. Squaring and adding gives (x − 4)²/36 + (y + 1)²/9 = 1, an ellipse centered at (4, −1).
5. Which curve is x = 3 tan t, y = 2 sec t?
Explanation: tan t = x/3 and sec t = y/2. Using sec² t − tan² t = 1 gives y²/4 − x²/9 = 1, a vertical hyperbola.
Common Mistakes to Avoid
- Listing a point as (t, x) or swapping coordinates. A parametric point is always (x(t), y(t)), with x first.
- Forgetting the direction of motion. A parametric curve is traced in the order of increasing t, so add arrows and never connect points in the wrong order.
- Finding intercepts by solving the wrong equation. y-intercepts come from x(t) = 0 and x-intercepts from y(t) = 0; then substitute that t into the other function.
- Checking only the vertex of a quadratic when finding extrema. Always compare the endpoints of the domain too.
- Confusing the ellipse and hyperbola relationships: ellipse is c² = a² − b², hyperbola is c² = a² + b². For an ellipse a is the larger denominator, but for a hyperbola a goes with the positive term.
- Reading sign errors in standard form: (x + 3)² means h = −3 and (y + 1)² means k = −1.
- Using the denominator instead of its square root in a trigonometric parametrization: for x²/16, the multiplier on cos t is 4, not 16.
- Dropping the domain after eliminating the parameter. The rectangular equation often shows more of the curve than the original t-interval draws (like a full circle from a half-circle).
- Dividing by zero when finding slope from parametric rates: if Δx/Δt = 0 the slope is undefined (vertical), not 0.
- Assuming a parametrization is unique. Many different pairs of equations trace the same curve, so check yours by substituting, not by comparing with someone else’s answer.
Frequently Asked Questions
What is the difference between a parametric equation and a regular function?
A regular function y = f(x) gives y from x alone. Parametric equations give both x and y as functions of a third variable t, usually time. This lets the curve cross itself, move back and forth, or form a closed loop such as a circle, which a single function y = f(x) cannot do, and it tells you where the particle is at each moment.
How do I decide which way a parametric curve is traced?
Follow the points as t increases. You can also check whether x(t) and y(t) are increasing or decreasing: increasing x means moving right, increasing y means moving up, and so on.
How do I find the center and size of an ellipse from its equation?
Put the equation in standard form with 1 on the right. The numbers inside the squares give the center (h, k). The larger denominator is a² (along the major axis) and the smaller is b². Then c = √(a² − b²) locates the foci.
Why does a hyperbola use sec and tan?
Because sec² t − tan² t = 1 has the same form as the hyperbola equation x²/a² − y²/b² = 1, just as cos² t + sin² t = 1 matches the ellipse and the circle.
Is this topic on the AP Precalculus exam?
The chapter notes that conic sections and these parametrization topics are not assessed on the AP exam, but they may be part of your school’s curriculum. Parametric motion with vectors and matrices is covered in the next module, which the second quiz in this module practices.
Where can I practice more problems like these?
The AP Precalculus Parametric Functions, Vectors, and Matrices Quiz 1 in the AP Precalculus QBank gives you more practice on this topic, and you can browse every other topic in the full AP Precalculus QBank.