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Line tangent to a parabola Example

Question:

Line y = 2x + n is tangent to the parabola y2 = 8x at the point A. Find the equation of the tangent line and the point of tangency.

find equation of the tangent line and the point of tangency

Solution:

The intersection point of a line and a curve is the solution of the two equations.

name y = 2x + n as equation1
name y2 = 8x as equation2
square both sides of the equation1
y2 = (2x + n)2
= 4x2 + 4nx + n2
name y2 = 4x2 + 4nx + n2 as equation3
substitute equation3 into equation2
4x2 + 4nx + n2 = 8x
combine similar terms
4x2 + (4n - 8)x + n2 = 0

Now we get a one variable second degree equation. Because the line and the curve have only one intersection point, then this one variable second degree equation has two identical solutions. That is, this one variable second degree equation has two identical roots, then there is b2 - 4ac = 0.

Because our one variable second degree equation is 4x2 + (4n - 8)2 + n2 = 0, so a = 4, b = 4n - 8, and c = n2

then b2 - 4ac = (4n - 8)2 - 4(4)n2
= 16n2 - 64n + 64 - 16n2
= - 64n + 64

From b2 - 4ac = 0, we get -64n + 64 = 0. Therefore, n = 1. So, the equation of the tangent line is y = 2x + 1.

Now, we find the point of the tangency.

Substitute n = 1 into the one variable second degree equation

4x2 + (4 × 1 - 8)x + 12 = 0
4x2 - 4x + 1 = 0
a = 4, b = -4, and c = 1
x1,2 = [-b +- square root of (b2 - 4ac)]/2a
= [4 +- square root of (16 - 4(4)(1))]/[2(4)]
=4/8 = 1/2
So, x1 = x2 = 1/2

y = 2x + 1 = 2 × (1/2) + 1 = 2. So, the coordinate of the point A is (1/2, 2).

graph of a line tangent to a parabola example

In the figure shown, the blue curve is the graph of the parabola y2 = 8x. The line is tangent to the parabola. The line has the equation y = 2x + 1. The point of tangency is A (1/2, 2). For more details, please watch the video.