Question

What is the equation for a parabola with a vertex at (5,-1) and a focus at (5,-7)?

Answer 1

`y=-1/24x^2+5/12x-49/24`

Explanation:

`"note that the vertex and focus are on the vertical axis"`

`x=5`

`"the focus and directrix are equidistant from the vertex"`

`"that is "-1-(-7)=6`

`"equation of directrix is "y=-1+6toy=5`

`"for any point "(x,y)" on the parabola"`

`"the distance to the focus and directrix are equidistant"`

`"using the "color(blue)"distance formula"`

`sqrt((x-5)^2+(y+7)^2)=|y-5|`

`color(blue)"squaring both sides"`

`(x-5)^2+(y+7)^2=(y-5)^2`

`(y+7)^2-(y-5)^2=-(x-5)^2`

`cancel(y^2)+14y+49cancel(-y^2)+10y-25=-x^2+10x-25`

`rArr24y=-x^2+10x-49`

`rArry=-1/24x^2+5/12x-49/24`

Answer 2

`-24(y+1)=(x-5)^2`

Explanation:

Let's graph the points we have.

The vertex and focus have the same x-coordinate. This tells us that this parabola doesn't open sideways, which means we'll be using the equation `4p(y-k)=(x-h)^2`
(Here's a link if you want to see how this equation was derived!)

In the equation, `p` is the distance from the vertex to the focus or directrix; `k` is the y-coordinate of the vertex, and `h` is the x-coordinate of the vertex.

Since the vertex is already given, we can plug that in:
`4p(y-k)=(x-h)^2`
`4p(y+1)=(x-5)^2`

To find the p-value, simply subtract the y-value of the vertex from the y-value of the focus.
`-7+1 = -6`
The p-value is negative. This makes sense, because the focus of the parabola is below the vertex, making the parabola open downwards. Downward parabolas have negative p-values.

Now, just plug that into the equation:
`4(-6)(y+1)=(x-5)^2`
`-24(y+1)=(x-5)^2`