 Finding the Intersections of Lines and Parabolas - dummies

# Finding the Intersections of Lines and Parabolas

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.