Question

How do you find the equation of the parabola described with Vertex at (2, -3); focus at (2, -5)?

Answer 1

`y=-1/8x^2+1/2x-7/2`

Explanation:

The axis passes through the vertex and the focus and its equation is

x=2

In the standard form of the parabola

`y=ax^2+bx+c`

the equation of the axis is `x=-b/(2a)`

so the first information you have is

1)`-b/(2a)=2`

The coordinates of the vertex and the focus are, respectively:

`V(-b/(2a),(4ac-b^2)/(4a))`

`F(-b/(2a),(1+4ac-b^2)/(4a))`

so you have the relations:

2)`(4ac-b^2)/(4a)=-3`

and

3)`(1+4ac-b^2)/(4a)=-5`

By (1) you have:

`b=-4a` that you can substitute in (2) and (3):

`(4ac-(-4a)^2)/(4a)=-3`

`(1+4ac-(-4a)^2)/(4a)=-5`

1) `b=color(red)(-4a)` that you can substitute in (2) and (3):
2) `(4ac-color(red)((-4a))^2)/(4a)=-3`
3) `(1+4ac-color(red)((-4a))^2)/(4a)=-5`

1) `b=-4a`
2) `4ac-16a^2=-12a`
3) `1+4ac-16a^2=-20a`

Then let's substitute the expression `4ac-16a^2` from (2) in (3):

1) `b=-4a`
2) `4ac-16a^2=color(green)(-12a)`
3) `1color(green)(-12a)=-20a`

Let's choose to solve (3):

3) `20a-12a=-1`
3) `a=-1/8`

and substitute in (1) to know b:

1) `b=-4(-1/8)=1/2`

Let's find c by substituting in (2) the value of a:

2) `4(-1/8)c-16(-1/8)^2=-12(-1/8)`

2)`-1/2c-16*1/64=3/2`

2)`-1/2c=1/4+3/2`

2)`c=-2(7/4)=-7/2`

Then the equation of the parabola is:

`y=-1/8x^2+1/2x-7/2`