Question

How do you find the vertex, focus and directrix of `x^2 + 6x + 8y + 25 = 0`?

Answer 1

When given the equation of a parabola, `y(x)=ax^2+bx+c`:
`h=-b/(2a)`
`k=y(h)`
`f=1/(4a)`
The vertex is: `(h,k)`
The focus is: `(h,k+f)`
The equation of the directrix: `y =k-f`

Explanation:

Write the given equation in the form, `y(x) = ax^2+bx+c`:

`x^2+6x+25+8y=0`

`x^2+6x+25=-8y`

`y(x) = -1/8x^2 -3/4x-25/8`

`a = -1/8, b = -3/4, and c = -25/8`

Compute the value of h:

`h = -b/(2a)`

`h = -(-3/4)/(2(-1/8))`

`h = -3`

Compute the value of k:

`k = y(h)`

`k = y(-3)`

`k = -1/8(-3)^2 - 3/4(-3)-25/8`

`k = -2`

Compute the value of f:

`f = 1/(4a)`

`f = 1/(4(-1/8))`

`f = -2`

The vertex is: `(-3,-2)`
The focus is: `(-3,-4)`
The equation of the directrix is `y = 0`