Conic sections

The curves obtained by slicing a cone with a plane that does not pass through the vertex are called conic sections or simply conic.  If the slicing plane is parallel to a generator, the cone is called parabola.



Standard form of a parabola

The standard form of a parabola is given by  y = ax2 + bx + c, a ≠ 0. It is the general quadratic equation.
If a > 0, the parabola opens upwards and if a < 0, the parabola opens downwards.

Vertex form of a parabola

The vertex form of a parabola is given by  y = a(x - h)2 + k, a ≠ 0.
If a > 0, the parabola opens upwards and if a < 0, the parabola opens downwards.
The vertex of the parabola is given by (h, k) and the axis of symmetry is given by x = h.

Lets see how to convert standard form to vertex form
To convert standard form to vertex form we follow the following procedure: -
  • First we have to take the common factors from the first two terms in the standard form f(x) = ax2 + bx + c
  • Now we complete the square of the first two terms by adding and subtracting square of the half of second term.
  • Simplify and rearrange it in the form y = a(x - h)2 + k

Lets consider a few examples.

Example 1: -
Convert the parabola f(x) = x2 - 2x + 3 in the standard form to vertex form and hence find the vertex.
Solution: -
Given standard form of parabola is f(x) = x2 - 2x + 3.
Comparing with the standard form of parabola, f(x) = ax2 + bx + c, we get
a = 1.
Since a = 1, we complete the square of the first two terms in the quadratic equation.  To complete the square, we add and subtract the square of the half of the second term.
So we get
f(x) = x2 - 2x + 1 - 1 + 3
Now we can combine the first three terms and last two term.
We get,
f(x) = (x - 1)2 + 2, which is the required vertex  form and the vertex is given by (1, 2).


Example 2: -
Convert the parabola f(x) = -4x2 + 3x  in the standard form to vertex form and hence find the vertex.
Solution: -
Given standard form of parabola is f(x) = -4x2 + 3x.
Comparing with the standard form of parabola, f(x) = ax2 + bx + c, we get
a = -4.
Since a = -4, we first take out -4 common from the first two terms
We get, f(x) = -4(x2 - 3/4x )
Now we complete the square of the first two terms in the quadratic equation.  To complete the square, we add and subtract the square of the half of the second term.
So we get
f(x) = -4(x2 - 3/4x + 9/64 - 9/64)
       = -4(x2 - 3/4x + 9/64) + 96
Now we can combine the first three terms and last two term.
We get,
f(x) = -4(x - 3/8)2 + 96, which is the required vertex form.  The vertex is given by (3/8, 96)
Example 3: -
Convert the parabola f(x) = -16x2 + 24x + 32  in the standard form to vertex form and hence find the vertex.
Solution: -
Given standard form of parabola is f(x) = -16x2 + 24x + 32
Comparing with the standard form of parabola, f(x) = ax2 + bx + c, we get
a = -16.
Since a = -16, we first take out -16 common from the first two terms
We get, f(x) = -16(x2 - 3/2x ) + 32
Now we complete the square of the first two terms in the quadratic equation.  To complete the square, we add and subtract the square of the half of the second term.
So we get
f(x) = -16(x2 - 3/2x + 9/4 - 9/4) + 32
       = -16(x2 - 3/2x + 9/16) + 9 + 32
Now we can combine the first three terms and last two term.
We get,
f(x) = -16(x - 3/4)2 + 41, which is the required vertex form.  The vertex is given by (3/4, 41)

Try yourself: -
Convert the parabola f(x) = x2 - 6x + 13  in the standard form to vertex form and hence find the vertex.

Solution: -
The vertex form is f(x) = (x - 3)2 + 4.  The vertex is (3, 4)