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Lesson Explainer: Points of Intersection of Parametric Equations Mathematics

In this explainer, we will learn how to find the points of intersection between parametric equations and given Cartesian equations.

Recall that a curve in the plane can be given by a pair of parametric equations 𝑥=𝑓(𝑡), 𝑦=𝑔(𝑡), which specify the coordinates of points on the curve in terms of a parameter 𝑡. Parameter 𝑡 can take any value in some specified range, often an interval, say, 𝑡∈[0,1] or 𝑡∈[0,2𝜋[, or the whole real line 𝑡∈ℝ.

We are going to see how to calculate the coordinates of points of intersection between curves given parametrically and lines specified by Cartesian equations. Conversely, we will see how to work out properties of parametric curves and properties of straight lines, given specified points of intersection.

Each problem has its own features, but the general approach is to use the given information to write down some equation in parameter 𝑡 and then solve this equation to find the value(s) of 𝑡 that satisfies the given conditions within the range of possible values for parameter 𝑡. We can then use this value of 𝑡 in the parametric equations of the curve to find the coordinates of the relevant points.

First, let us use a given 𝑥-intercept point to calculate the value of an unknown constant in a pair of parametric equations.

Example 1: Finding a Constant in a Parametric Equation given a Point the Curve Passes Through

The graph shows a curve given by the parametric equations 𝑥=𝑘𝑡+𝑡 and 𝑦=𝑘𝑡+27, where 𝑡∈ℝ and 𝑘≠0 is some constant. Given that the point (−6,0) lies on the curve, find the value of 𝑘.

Answer

Since the point (−6,0) lies on the curve, we can substitute the values 𝑥=−6 and 𝑦=0 into the parametric equations: −6=𝑘𝑡+𝑡,0=𝑘𝑡+27.

Dividing the second of these equations by the nonzero constant 𝑘 yields the equation 0=𝑡+27, which we can solve for 𝑡: 𝑡=−27𝑡=−3, telling us that the curve has a 𝑦-coordinate of 0 when parameter 𝑡=−3. Substituting this value of 𝑡 into the first equation −6=𝑘𝑡+𝑡, we have −6=𝑘×(−3)−3−3=9𝑘𝑘=−13.

Thus, the value of the constant 𝑘 is −13.

Now that we have calculated the value of the constant 𝑘, we have enough information to work out the coordinates of the curve’s 𝑦-intercepts. Any curve’s 𝑦-intercepts are the points on the curve with 𝑥-coordinate 0. Since a parametric curve is given by expressions 𝑥(𝑡) and 𝑦(𝑡) for the coordinates in terms of parameter 𝑡, solving the equation 𝑥(𝑡)=0 will give us the values of 𝑡 at which the curve crosses the 𝑦-axis.

Example 2: Finding the 𝑦-Axis Intersection of a Parametric Equation

The graph shows a curve given by parametric equations 𝑥=−13𝑡+𝑡 and 𝑦=−13𝑡+27, where 𝑡∈ℝ. Find the coordinates of the points 𝑃 and 𝑄 where the curve crosses the 𝑦-axis.

Answer

The curve crosses the 𝑦-axis at points with 𝑥-coordinate equal to 0. We therefore substitute 𝑥=0 into the parametric equation 𝑥=−13𝑡+𝑡 and rearrange: 0=−13𝑡+𝑡𝑡−3𝑡=0.

We can then factor to obtain 𝑡(𝑡−3)=0.

Therefore, the curve crosses the 𝑦-axis when 𝑡=0 and when 𝑡=3, which are both valid values of the parameter since 𝑡 ranges over all of ℝ. We substitute these values of the parameter into the equation 𝑦=−13𝑡+27 to find the 𝑦-coordinates of the points 𝑃 and 𝑄: 𝑦=−130+27=−9,𝑦=−133+27=−18.

Therefore, the curve crosses the 𝑦-axis at the points 𝑃=(0,−9) and 𝑄=(0,−18).

Suppose that we are given parametric equations 𝑥=𝑓(𝑡), 𝑦=𝑔(𝑡) of a curve and a linear Cartesian equation 𝑎𝑥+𝑏𝑦+𝑐=0 (𝑎, 𝑏, and 𝑐 are constants) and asked to find the point(s) of intersection. We can substitute the parametric equations for 𝑥 and 𝑦 into the Cartesian equation: 𝑎𝑓(𝑡)+𝑏𝑔(𝑡)+𝑐=0.

This will give us a single equation in 𝑡 that we can attempt to solve to find the value(s) of the parameter at the point(s) where the line and the curve meet. Substituting the value(s) back into the equations 𝑥=𝑓(𝑡) and 𝑦=𝑔(𝑡) will give us the coordinates of the intersection point(s).

Example 3: Finding the Coordinates of the Intersection between a Parametric Equation and a Cartesian Equation

Find the coordinates of the point where the parametric curve 𝑥=𝑡, 𝑦=8𝑡, where 𝑡∈ℝ, and the line 𝑦−4𝑥=4 meet.

Answer

The parametric equations of the curve give us the parametric expressions 𝑥=𝑡, 𝑦=8𝑡 for the coordinates of the points on the curve. At the point where the curve meets the line, these coordinates must also satisfy the Cartesian equation 𝑦−4𝑥=4. We therefore substitute 𝑥=𝑡, 𝑦=8𝑡 into 𝑦−4𝑥=4: 8𝑡−4𝑡=4. Then, we solve for 𝑡: 4𝑡−8𝑡+4=0𝑡−2𝑡+1=0(𝑡−1)=0𝑡=1.

This shows that, at the point where the line and the curve meet, parameter 𝑡=1, which is valid since 𝑡 can take any real value. We substitute this value of 𝑡 back into the parametric equations of the curve to find the coordinates of the point of intersection: 𝑥=𝑡=1=1,𝑦=8𝑡=8×1=8.

The coordinates of the point where the line and the curve meet are therefore (1,8).

Suppose that we are given parametric equations 𝑥=𝑓(𝑡), 𝑦=𝑔(𝑡) of a curve and the equation of a horizontal line 𝑦=𝑎 (𝑎 is a constant) or a vertical line 𝑥=𝑏 (𝑏 is a constant). In this case, we can directly set the relevant parametric coordinate equation equal to the constant: either 𝑓(𝑡)=𝑏 or 𝑔(𝑡)=𝑎. In both cases, we have a single equation that we can solve for 𝑡 as before.

Example 4: Finding the Coordinates of the Intersection between a Parametric Equation and a Cartesian Equation

The graph shows the parametric curve 𝑥=(𝑡)−(𝑡)sincos, 𝑦=𝑡+𝜋2, where −𝜋<𝑡<𝜋. Find the coordinates of the point where the curve meets the line 𝑦=𝜋.

Answer

We can substitute the equation of the line 𝑦=𝜋 directly into the parametric equation 𝑦=𝑡+𝜋2 in order to find 𝑡: 𝑡+𝜋2=𝜋𝑡+𝜋2=±𝜋.

Therefore, we have 𝑡=𝜋−𝜋2=𝜋2 or 𝑡=−𝜋−𝜋2=−3𝜋2. Note, however, that −3𝜋2 is outside of the specified range −𝜋<𝑡<𝜋. Therefore, 𝑡=𝜋2. Since we are given the 𝑦-coordinate 𝑦=𝜋, we just need to substitute 𝑡=𝜋2 into the parametric equation 𝑥=(𝑡)−(𝑡)sincos to find the 𝑥-coordinate: 𝑥=𝜋2−𝜋2=1−0=1.sincos

Hence, the parametric curve 𝑥=(𝑡)−(𝑡)sincos, 𝑦=𝑡+𝜋2 meets the line 𝑦=𝜋 at the point 1,𝜋, for −𝜋<𝑡<𝜋.

Suppose that we are given the parametric equations of a curve 𝑥=𝑓(𝑡), 𝑦=𝑔(𝑡) and told that this curve intersects a straight line at specified values of parameter 𝑡=𝑡 and 𝑡=𝑡. We can evaluate the points (𝑓(𝑡),𝑔(𝑡)) and (𝑓(𝑡),𝑔(𝑡)), giving us the coordinates of the points of intersection, (𝑥,𝑦) and (𝑥,𝑦).

From the coordinates of these two points, we can calculate an equation for the line that passes through them using the standard formula 𝑦−𝑦=𝑦−𝑦𝑥−𝑥(𝑥−𝑥), as in the next example.

Example 5: Finding the Cartesian Equation of a Line given the Points Where It Intersects a Parametric Curve

A curve is defined by the parametric equations 𝑥=(𝑡)cos, 𝑦=(𝑡)sin, where 0≤𝑡<2𝜋. A straight line 𝐿 intersects the curve at the points where 𝑡=𝜋3 and 𝑡=𝜋. Write the equation of line 𝐿 in the form 𝑦=𝑚𝑥+𝑏.

Answer

We first recall that the equations 𝑥=(𝑡)cos and 𝑦=(𝑡)sin are the parametric equations of a circle of radius 1 centered at the origin. The values 𝑡=𝜋3 and 𝑡=𝜋 give us two points on the circle; we need to find the equation of the straight line passing through these two points.

Substituting in the given values of 𝑡, the two points on the circle have coordinates 𝜋3,𝜋3=12,√32,((𝜋),(𝜋))=(−1,0).cossincossin

Recall that given two points (𝑥,𝑦) and (𝑥,𝑦) in the plane, we can write an equation for the line passing through them as 𝑦−𝑦=𝑦−𝑦𝑥−𝑥(𝑥−𝑥), and so we have 𝑦−√32=⎛⎜⎜⎝0−−1−⎞⎟⎟⎠𝑥−12𝑦=√33𝑥−12+√32𝑦=√33𝑥+√33.√

Let us consider now the question of tangency to parametric curves. We are going to look at a technique for judging whether a particular line is tangent to a particular curve or not.

Suppose we are given parametric equations of a curve 𝑥=𝑓(𝑡), 𝑦=𝑔(𝑡) and a line 𝑦=𝑚𝑥+𝑏. The first condition for a line to be tangent to a curve at a point 𝑝=(𝑓(𝑡),𝑔(𝑡)) is that the line and the curve intersect at that point. That is, the coordinates of the point must satisfy the equation of the line: 𝑔(𝑡)=𝑚𝑓(𝑡)+𝑏.

If the parametric expressions 𝑓(𝑡) and 𝑔(𝑡) are polynomials, then the equation 𝑔(𝑡)=𝑚𝑓(𝑡)+𝑏 will also be a polynomial. If 𝑓(𝑡) and 𝑔(𝑡) are given in terms of trigonometric functions, then we may be able to use trigonometric identities such as the Pythagorean identity sincos(𝜃)+(𝜃)=1 to reexpress 𝑔(𝑡)=𝑚𝑓(𝑡)+𝑏 as a polynomial.

The point here is that if 𝑔(𝑡)=𝑚𝑓(𝑡)+𝑏 is a polynomial, then it has a discriminant. The condition that this discriminant is zero corresponds to the polynomial having a repeated root, which in turn means that the line and the curve are tangent at the point.

If the given parametric curve is sufficiently simple, then the polynomial equation we get will be a quadratic: 𝐴𝑡+𝐵𝑡+𝐶=0. In this case, the discriminant is Δ=𝐵−4𝐴𝐶 and the condition of tangency is that 𝐵−4𝐴𝐶=0.

A positive discriminant here means that the line intersects the curve but is not a tangent, while a negative discriminant means that the line and the curve are disjoint.

Note that if the given parametric curve is more complicated, then the polynomial may not be a quadratic: it could be cubic or even have a higher degree. If this is the case, then the discriminant can still be used to check tangency, but it is more difficult and beyond the scope of this explainer.

In summary, the procedure for checking whether a given line is tangent to a parametric curve is as follows.

How To: Checking Whether a Line Is Tangent to a Parametric Curve

Suppose we are given parametric equations of a curve 𝑥=𝑓(𝑡), 𝑦=𝑔(𝑡) and a line 𝑦=𝑚𝑥+𝑏.

  1. Substitute the parametric expressions 𝑥=𝑓(𝑡), 𝑦=𝑔(𝑡) into the equation of the line: 𝑔(𝑡)=𝑚𝑓(𝑡)+𝑏.
  2. Reexpress the equation 𝑔(𝑡)=𝑚𝑓(𝑡)+𝑏 as a polynomial 𝑃(𝑡)=0, using trigonometric identities if necessary.
  3. In sufficiently simple cases, 𝑃(𝑡) will be a quadratic: 𝑃(𝑡)=𝐴𝑡+𝐵𝑡+𝐶. We can check for tangency using the discriminant Δ=𝐵−4𝐴𝐶 of this quadratic: Δ=0∶.Δ>0∶.Δ<0∶.ThelineandthecurvearetangentThelineandthecurveintersectintwopointsThelineandthecurvearedisjoint

In the next example, we will see how to find equations of tangent lines when their gradients are given.

Example 6: Finding Tangents to a Parametric Curve

A parametric curve is given by the equations 𝑥=𝑡+1 and 𝑦=4𝑡𝑡+2, where 𝑡≠−2. Find the equations of the two tangents to the curve that are parallel to the straight line passing through the 𝑥- and 𝑦-intercepts of the curve. Give your answers in the form 𝑦=𝑚𝑥+𝑏.

Answer

We are looking for tangent lines to the curve given by 𝑥=𝑡+1, 𝑦=4𝑡𝑡+2, where 𝑡≠−2, that are parallel to the line that passes through the 𝑥- and 𝑦-intercepts of the curve. Let us start by finding the slope of this line.

The 𝑥-intercepts of the curve are the points on the curve with 𝑦-coordinate equal to zero. Looking at the parametric equation 𝑦=4𝑡𝑡+2, we can see that the 𝑦-coordinate is equal to zero when the numerator 4𝑡=0, which happens when 𝑡=0. When 𝑡=0, we have 𝑥=𝑡+1=0+1=1. Thus, the 𝑥-intercept of the curve has coordinates (1,0). On the other hand, the 𝑦-intercept has 𝑥=𝑡+1=0, which holds when 𝑡=−1, giving 𝑦=4𝑡𝑡+2=−4−1+2=−4. So, the 𝑦-intercept of the curve has coordinates (0,−4).

Let us sketch the given parametric curve, marking the intercepts that we have just calculated.

The equation of the line passing through the points (0,−4) and (1,0) is 𝑦=4𝑥−4.

We are looking for lines parallel to this line, that is, lines with the equation 𝑦=4𝑥+𝑏, for some 𝑏. The other condition on these lines is that they should be tangent to the given parametric curve. One thing that we know about tangents to a curve is that they intersect the curve. That is, there must be points (𝑥,𝑦)=𝑡+1,4𝑡𝑡+2, where 𝑡≠2, on the curve that satisfy the equation 𝑦=4𝑥+𝑏. We can express this condition by substituting our parametric expressions for 𝑥 and 𝑦 into the equation of the line. This yields the equation 4𝑡𝑡+2=4(𝑡+1)+𝑏.

Since 𝑡 does not take the value −2, we can multiply everything by 𝑡+2: 4𝑡=(𝑡+2)(4(𝑡+1)+𝑏)=4𝑡+4𝑡+𝑏𝑡+8𝑡+8+2𝑏.

This equation simplifies to the following quadratic equation in 𝑡: 4𝑡+(𝑏+8)𝑡+8+2𝑏=0.

The solutions to this equation are the values of 𝑡 at which the parametric curve intersects the line 𝑦=4𝑥+𝑏. However, we are not looking for lines that simply intersect the curve; we want them to be tangent. Tangents to a degree 2 curve such as this one intersect the curve at precisely one point. This corresponds to values of 𝑏 for which the quadratic 4𝑡+(𝑏+8)𝑡+8+2𝑏=0 has exactly one real root. Recall that a quadratic has a single real root if and only if its discriminant Δ=𝐵−4𝐴𝐶 is equal to zero. Thus, we are looking for values of 𝑏 satisfying (𝑏+8)−4×4×(8+2𝑏)=0.

This simplifies to another quadratic, this time in 𝑏: 𝑏−16𝑏−64=0.

A quick application of the quadratic formula yields the solutions 𝑏=8+8√2 and 𝑏=8−8√2. Thus, we have a pair of tangent lines: 𝑦=4𝑥+8+8√2 and 𝑦=4𝑥+8−8√2.

Let us finish by recapping a few important concepts from this explainer.

Key Points

  • Given parametric equations of a curve 𝑥=𝑓(𝑡), 𝑦=𝑔(𝑡) and a linear Cartesian equation 𝑎𝑥+𝑏𝑦+𝑐=0, we can write a single equation 𝑎𝑓(𝑡)+𝑏𝑔(𝑡)+𝑐=0, which can be solved to find the value(s) of 𝑡 when the curve intersects the line. We can then work out the coordinates of the point(s) of intersection of the curve and the line.
  • Given parametric equations of a curve 𝑥=𝑓(𝑡), 𝑦=𝑔(𝑡) and the equation of a horizontal line 𝑦=𝑎 or a vertical line 𝑥=𝑏, we can write either 𝑓(𝑡)=𝑏 or 𝑔(𝑡)=𝑎. Solving this equation for 𝑡 gives the value(s) of the parameter at the point(s) of intersection. This technique works in particular for finding 𝑥- and 𝑦-intercepts of parametric curves, in which case the relevant lines are 𝑦=0 and 𝑥=0.
  • Given values of parameter 𝑡=𝑇 allow us to calculate the coordinates of points on the curve (𝑓(𝑇),𝑔(𝑇)). We can use standard methods to write equations of straight lines passing through these points.
  • We can check whether sufficiently simple parametric curves and Cartesian straight lines intersect, are tangential, or do not meet by looking at the discriminant of the equation that results from substituting the parametric expressions for 𝑥 and 𝑦 into the equation of the line:
    • When the discriminant Δ=𝐵−4𝐴𝐶=0, the line and the curve are tangent.
    • When Δ>0, the line and the curve intersect but are not tangent.
    • When Δ<0, the line and the curve are disjoint.

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