Solved Examples: Finding the Coordinates of Focus of a Parabola

Solved Example 1

`y = x^2`

Solution

Step 1: Convert the given equation to vertex form `y = a(x - h)^2 + k`

The given equation is already in the vertex form as it can be written as `y = 1(x - 0)^2 + 0`.

Step 2: Find 'a', 'h', and 'k'.

On comparing `y = 1(x - 0)^2 + 0` with `y = a(x - h)^2 + k` we get,

`a = 1`

`h = 0`

`k = 0`

Step 3: The coordinates of the focus are given by `(h, k + 1/(4a))`

Plug in the values of 'a', 'h' and 'k' to get

`(0, 0 + 1/(4*1))`

Simplifying,

`(0, 1/4)`

Thus `(0, 1/4)` is the focus of the parabola `y = x^2`.

Solved Example 2

`y = 4x^2`

Solution

Step 1: Convert the given equation to vertex form `y = a(x - h)^2 + k`

The given equation is already in the vertex form as it can be written as `y = 4(x - 0)^2 + 0`.

Step 2: Find 'a', 'h', and 'k'.

On comparing `y = 4(x - 0)^2 + 0` with `y = a(x - h)^2 + k` we get,

`a = 4`

`h = 0`

`k = 0`

Step 3: The coordinates of the focus are given by `(h, k + 1/(4a))`

Plug in the values of 'a', 'h' and 'k' to get

`(0, 0 + 1/(4*4))`

Simplifying,

`(0, 1/(16))`

Thus `(0, 1/(16))` is the focus of the parabola `y = 4x^2`.

Solved Example 3

`y = 5x^2 + 10`

Solution

Step 1: Convert the given equation to vertex form `y = a(x - h)^2 + k`

The given equation is already in the vertex form as it can be written as `y = 5(x - 0)^2 + 10`.

Step 2: Find 'a', 'h', and 'k'.

On comparing `y = 5(x - 0)^2 + 10` with `y = a(x - h)^2 + k` we get,

`a = 5`

`h = 0`

`k = 10`

Step 3: The coordinates of the focus are given by `(h, k + 1/(4a))`

Plug in the values of 'a', 'h' and 'k' to get

`(0, 10 + 1/(4*5))`

Simplifying,

`(0, 201/20)`

Thus `(0, 201/20)` is the focus of the parabola `y = 5x^2 + 10`.

Solved Example 4

`y = 4x^2 + 2x + 2`

Solution

Step 1: Convert the given equation to vertex form `y = a(x - h)^2 + k`

Since this equation is not already in the vertex form (unlike the examples above), we'll use the method of completing the square to convert this equation to the vertex form. First, factor out 4 from the LHS,

`y = 4(x^2 + x/2 + 1/2)`

Now, identify the coefficient of 'x', which is `1/2`. Divide it by 2 to get `1/4`. Square it to get `(1/4)^2`. Add and subtract this number from the quadratic expression in the parenthesis,

`y = 4(x^2 + x/2 + (1/4)^2 - (1/4)^2 + 1/2)`

Compare the highlighted part of the equation above with `a^2 + 2ab + b^2` to get `a = x` and `b = 1/4`. Now apply the formula `a^2 + 2ab + b^2 = (a + b)^2`,

`y = 4((x + 1/4)^2 - (1/4)^2 + 1/2)`

Simplify the LHS,

 `y = 4((x + 1/4)^2 + 7/16)`

The above equation is now in the vertex form.

Step 2: Find 'a', 'h', and 'k'.

On comparing `y = 4((x + 1/4)^2 + 7/16)` with `y = a(x - h)^2 + k` we get,

`a = 4`

`h = -1/4`

`k = 7/16`

Step 3: The coordinates of the focus are given by `(h, k + 1/(4a))`

Plug in the values of 'a', 'h' and 'k' to get

`(-1/4, 7/16 + 1/(4*4))`

Simplifying,

`(-1/4, 1/2)`

Thus `(-1/4, 1/2)` is the focus of the parabola `y = 4x^2 + 2x + 2`.

Solved Example 5

`y = (x - 1)^2`

Solution

Step 1: Convert the given equation to vertex form `y = a(x - h)^2 + k`

The given equation is already in the vertex form as it can be written as `y = 1(x - 1)^2 + 0`.

Step 2: Find 'a', 'h', and 'k'.

On comparing `y = 1(x - 1)^2 + 0` with `y = a(x - h)^2 + k` we get,

`a = 1`

`h = 1`

`k = 0`

Step 3: The coordinates of the focus are given by `(h, k + 1/(4a))`

Plug in the values of 'a', 'h' and 'k' to get

`(1, 0 + 1/(4*1))`

Simplifying,

`(1, 1/4)`

Thus `(1, 1/4)` is the focus of the parabola `y = (x - 1)^2`.

Solved Example 6

`y = 2(x + 5)^2 + 6`

Solution

Step 1: Convert the given equation to vertex form `y = a(x - h)^2 + k`

The given equation is already in the vertex form.

Step 2: Find 'a', 'h', and 'k'.

On comparing `y = 2(x + 5)^2 + 6` with `y = a(x - h)^2 + k` we get,

`a = 2`

`h = -5`

`k = 6`

Step 3: The coordinates of the focus are given by `(h, k + 1/(4a))`

Plug in the values of 'a', 'h' and 'k' to get

`(-5, 6 + 1/(4*2))`

Simplifying,

`(-5, 49/8)`

Thus `(-5, 49/8)` is the focus of the parabola `y = 2(x + 5)^2 + 6`.

Solved Example 7

`y = (x + 1)(x + 2)`

Solution

Step 1: Convert the given equation to vertex form `y = a(x - h)^2 + k`

The given equation is in the intercept form. We need to first expand it to the standard form by using FOIL.

`y = x^2+ 3x + 2`

Now we will apply the method of completing the square to convert it to vertex form.

Take the coefficient of `x`, which is 3, divide it by 2 to get `3/2` and then square it to get `(3/2)^2`. Add and subtract `(3/2)^2` from the LHS,

`y = x^2 + 3x + (3/2)^2 - (3/2)^2 + 2`

Compare the highlighted part of the equation above with `a^2 + 2ab + b^2` to get `a = x` and `b = 3/2`. Now apply the formula `a^2 + 2ab + b^2 = (a + b)^2`,

`y = (x + 3/2)^2 - (3/2)^2 + 2`

Simplify the LHS,

`y = (x + 3/2)^2 - 1/4`

The above equation is now in the vertex form.

Step 2: Find 'a', 'h', and 'k'.

On comparing `y = (x + 3/2)^2 - 1/4` with `y = a(x - h)^2 + k` we get,

`a = 1`

`h = -3/2`

`k = -1/4`

Step 3: The coordinates of the focus are given by `(h, k + 1/(4a))`

Plug in the values of 'a', 'h' and 'k' to get

`(-3/2, -1/4 + 1/(4*1))`

Simplifying,

`(-3/2, 0)`

Thus `(-3/2, 0)` is the focus of the parabola `y = (x + 1)(x + 2)`.

Solved Example 8

`x = 4y^2`

Solution

The above equation is different from earlier equations discussed above in that it is not a function. The y-term is squared while the x-term is the dependent variable. It's graph does not pass the vertical line test. Although it is not a function, it can be graphed on a coordinate plane and its focus can be calculated by the formula `(k + 1/(4a), h)`

Step 1: Convert the given equation to vertex form `x = a(y - h)^2 + k`

The given equation is already in the vertex form as it can be written as `x = 4(y - 0)^2 + 0`.

Step 2: Find 'a', 'h', and 'k'.

On comparing `x = 4(y - 0)^2 + 0` with `y = a(x - h)^2 + k` we get,

`a = 4`

`h = 0`

`k = 0`

Step 3: The coordinates of the focus are given by `(k + 1/(4a), h)`

Plug in the values of 'a', 'h' and 'k' to get

`(0 + 1/(4*1), 0)`

Simplifying,

`(1/16, 0)`

Thus `(1/16, 0)` is the focus of the parabola `x = 4y^2`.

Solved Example 9

`x = 5y^2 - 1`

Solution

This equation is same as the above one in that it is not a function but only a quadratic relation.

Step 1: Convert the given equation to vertex form `y = a(x - h)^2 + k`

The given equation is already in the vertex form as it can be written as `x = 5(y - 0)^2 - 1`.

Step 2: Find 'a', 'h', and 'k'.

On comparing `x = 5(y - 0)^2 - 1` with `y = a(x - h)^2 + k` we get,

`a = 5`

`h = 0`

`k = -1`

Step 3: The coordinates of the focus are given by `(k + 1/(4a), h)`

Plug in the values of 'a', 'h' and 'k' to get

`(-1 + 1/(4*5), 0)`

Simplifying,

`(-19/20, 0)`

Thus `(-19/20, 0)` is the focus of the parabola `x = 5y^2 - 1`.

Solved Example 10

`x = 6y^2 +5y - 1`

Solution

Step 1: Convert the given equation to vertex form `y = a(x - h)^2 + k`

We will apply the method of completing the square to convert it to vertex form. First, factor out 6 from the LHS.

`x =6(y^2 + 5/6y -1/6)`

Take the coefficient of `y`, which is `5/6`, divide it by 2 to get `5/12` and then square it to get `(5/12)^2`. Add and subtract `(5/12)^2` from the LHS,

`x =6( Compare the highlighted part of the equation above with `a^2 + 2ab + b^2` to get `a = y` and `b = 5/12`. Now apply the formula `a^2 + 2ab + b^2 = (a + b)^2`,

`x = 6((y + 5/12)^2 - (5/12)^2 - 1/6)`

Simplify the LHS,

`x = 6(y + 5/12)^2 - 31/6`

The above equation is now in the vertex form.

Step 2: Find 'a', 'h', and 'k'.

On comparing `x = 6(y + 5/12)^2 - 31/6` with `y = a(x - h)^2 + k` we get,

`a = 6`

`h = -5/12`

`k = -31/6`

Step 3: The coordinates of the focus are given by `(k + 1/(4a), h)`

Plug in the values of 'a', 'h' and 'k' to get

`(-31/6 + 1/(4*6), -5/12)`

Simplifying,

`(-41/8, -5/12)`

Thus `(-41/8, -5/12)` is the focus of the parabola `x = 6y^2 +5y - 1`.