Intersection of Lines and Parabolas

Video lesson · from 5:092 worked examples

One lesson video covers all of Polynomials & Quadratics, so it opens at 5:09 for this topic — not from the beginning.

A line may:

If we have a line with equation of the form $y=mx+k$ and parabola with equation of the form $y=px^2+qx+r$ , equating gives us:

mx+k=px^2+qx+r

px^2+qx-mx+r-k=0

px^2+(q-m)x+(r-k)=0

This gives us a quadratic equation; hence, we can conclude:

b² − 4ac > 0

The line meets the
parabola at two
distinct points

b² − 4ac = 0

The line meets the
parabola once
i.e. the line is a tangent
to the parabola

b² − 4ac < 0

The line does not
meet the parabola

State the condition you are testing: two intersections need $b^2-4ac \gt 0$ , a tangent needs $= 0$ , and no intersection needs $\lt 0$ . Write it down before you solve — leaving it unstated is one of the most repeated faults on these questions.
Rearrange to zero first: substitute the line into the curve, then collect everything on one side before reading off $a$ , $b$ and $c$ .
Use brackets when substituting: especially with a negative $b$ or a coefficient in terms of $k$ .
"Tangent" and "equal roots" are the same statement: and both mean the discriminant is zero.

Example 1

Find the coordinates of the points where the line with equation $y=2x+1$ intersects the curve with equation $y=x^2-2$ .

Equate the expressions for $y$ :

x^2 - 2 = 2x + 1

Rearrange into standard quadratic form ( $ax^2+bx+c=0$ ):

x^2 - 2x - 3 = 0

Factorise to solve for $x$ :

(x - 3)(x + 1) = 0

So $x = 3$ or $x = -1$ .

Substitute these $x$ values back into the linear equation (it's usually simpler) to find the corresponding $y$ values:

If $x = 3$ , $y = 2(3) + 1 = 7$ . (Point: $(3, 7)$ )

If $x = -1$ , $y = 2(-1) + 1 = -1$ . (Point: $(-1, -1)$ )

The points of intersection are $(3, 7)$ and $(-1, -1)$ .

Example 2

Show that the line with equation $y=5x-2$ is a tangent to the parabola $y=2x^2+x$ and find the point of contact.

Equate the expressions for $y$ :

2x^2 + x = 5x - 2

2x^2 - 4x + 2 = 0

Check the discriminant $b^2 - 4ac$ of this resulting quadratic, where $a=2, b=-4, c=2$ :

b^2 - 4ac = (-4)^2 - 4(2)(2) = 16 - 16 = 0

Since $b^2 - 4ac = 0$ , there is 1 real and equal root. The line meets the parabola exactly once, so it is a tangent.

To find the point of contact, solve the equation:

2(x^2 - 2x + 1) = 0

x^2 - 2x + 1 = 0

(x - 1)(x - 1) = 0

So $x = 1$ .

Substitute $x = 1$ into the linear equation to find $y$ :

y = 5(1) - 2 = 3

The point of contact is $(1, 3)$ .