Question

How do you write the parabola `y^2-10y-27x+133=0` in standard form and find the vertex, focus, and directrix?

Answer 1

Standard form: `(y-5)^2 = 4*27/4(x-4)`
Vertex is at ( 4,5) , Focus is at `(43/4), 5`. directrix is `y = -11/7`

Explanation:

`y^2-10y-27x+133=0 or y^2-10y+25-25-27x+133=0` or
`(y-5)^2 =27x -108 or (y-5)^2 = 27(x-4)` or
`(y-5)^2 = 4*27/4(x-4)`
Comparing with standard equation `(y-k)^2 = 4*a(x-h); (h,k)` being vertex. we get here `h = 4 , k = 5 , a= 27/4`

Vertex is at ( 4,5) , Focus is at `((h+a),k) or (4 + 27/4), 5 or (43/4,5)`.

Vertex is equidistant from focus and directrix.

So directrix is `y= (h-a) = 4-27/4 or y = -11/7`
graph{(y-5)^2=27(x-4) [-160, 160, -80, 80]} [Ans]

Answer 2

Vertex is `(4,5)` and focus is `(43/4,5)` and axis of symmetry is `x=-11/4`.

Explanation:

Standard form of equation of parabola is `y=ax^2+bx+c` or `x=ay^+by+c`. Asin the given equation, we have `y^2`, we can write `y^2-10y-27x+133=0` as

`27x=y^2-10y+133` and in standard form it is

`x=1/27y^2-10/27y+133/27`

Vertex form of equation of paarbola is `y=a(x-h)^2+k` or `x=(y-k)^2+h`, where `x=h` or `y=k` is axis of symmetry (focus and vertex lie on this and directrix is perpendicular to it) and vertex is `(h,k)`.

In the equation `y=a(x-h)^2+k`, focus is `(h,k+1/(4a))` and directrix is `y=k-1/(4a)`. In case equation is `x=(y-k)^2+h`, focus is `(h+1/(4a),k)` and directrix is `x=h-1/(4a)`.

`27x=y^2-10y+133=y^2-2xx5xxy+25-25+133`

`27x=(y-5)^2+108` or `x=1/27(y-5)^2+4`

This is the vertex form of this equation and vertex is `(4,5)`. As we have `a=1/27`, `1/(4a)=27/4` and focus is `(4+27/4,5)` i.e. `(43/4,5)` and axis of symmetry is `x=4-27/4=-11/4`.

graph{(y^2-10y-27x+133)(x+11/4)((x-43/4)^2+(y-5)^2-2)=0 [-45.3, 114.64, -35.3, 44.7]}