Finding the Intersections of Lines and Parabolas

By Mary Jane Sterling

A line can cut through a parabola in two points, or it may just be tangent to the parabola and touch it at one point. And then, sadly, a line and a parabola may never meet. When solving systems of equations involving lines and parabolas, you usually use the substitution method — solving for x or y in the equation of the line and substituting into the equation of the parabola.

Sometimes the equations lend themselves to elimination — when adding the equations (or multiples of the equations) together eliminates one of the variables entirely because its coefficient becomes 0. Elimination works only occasionally, but substitution always works.

Sample questions

  1. Find the common solution(s) in the equations y = –5x2 + 12x + 3 and 8x + y = 18.

    The points of intersection are (1, 10), (3, –6). Here’s another way to write this solution: When x = 1, y = 10, and when x = 3, y = –6. To find these solutions, rewrite the equation of the line as y = 18 – 8x.

    Replace the y in the equation of the parabola with its equivalent to get 18 – 8x = –5x2 + 12x + 3. Move all the terms to the left and combine like terms, giving you 5x2 – 20x + 15 = 0. Divide each term by 5 and then factor, which gives you the equation 5(x2 – 4x + 3) = 5(x – 3)(x – 1) = 0.

    Using the multiplication property of zero (in order for a product to equal 0, one of the factors must be 0), you know that x = 3 or x = 1. Substitute those values back into the equation of the line to get the corresponding y-values.

    Always substitute back into the equation with the lower exponents. You can avoid creating extraneous solutions.

  2. Find the common solution(s) in the equations y = x2 – 4x and 2x + y + 1 = 0

    (1, –3). Solve for y in the equation of the line to get y = –2x – 1. Substitute this value into the equation of the parabola to get –2x – 1 = x2 – 4x. Moving the terms to the right and simplifying, 0 = x2 – 2x + 1 = (x – 1)2.

    The only solution is x = 1. Replacing x with 1 in the equation of the line, you find that y = –3. The line is tangent to the parabola at the point of intersection, which is why this problem has only one solution.

Practice questions

  1. Find the common solution(s) in the equations y = x2 + 4x + 7 and 3xy + 9 = 0.

  2. Find the common solution(s) in the equations y = 4x2 – 8x – 3 and 4x + y = 5.

Following are answers to the practice questions:

  1. The answer is (–2, 3), (1, 12).

    Solve for y in the second equation (you get y = 3x + 9), and substitute that into the equation of the parabola: 3x + 9 = x2 + 4x + 7. Move all the terms to the right and factor the equation: 0 = x2 + x – 2 = (x + 2)(x – 1).

    So, x = –2 or 1. Letting x = –2 in the equation of the line, 3(–2) – y + 9 = 0; –6 – y + 9 = 0; –y = –3; y = 3. And when x = 1 in the equation of the line, 3(1) – y + 9 = 0; 3 – y + 9 = 0; –y = –12; y = 12.

    When solving for the second coordinate in the solution of a system of equations, use the simpler equation — the one with the smaller exponents — to avoid introducing extraneous solutions.

  2. The answer is (–1, 9), (2, –3).

    Solve for y in the second equation (you get y = 5 – 4x) and substitute the equivalent of y into the equation of the parabola: 5 – 4x = 4x2 – 8x – 3. Move all the terms to the right and factor the equation: 0 = 4x2 – 4x – 8 = 4(x2x – 2) = 4(x + 1)(x – 2).

    Using the multiplication property of zero, you find that x = –1 or x = 2. When x = –1 in the equation of the line, 4(–1) + y = 5; –4 + y = 5; y = 9. And substituting x = 2 in the equation of the line, 4(2) + y = 5; 8 + y = 5; y = –3.