Question

Conics Question: Graph the equation x=-3y^2+12y+13 What are all applicable points? (vertex, focus, etc.)

Answer 1

Vertex is `(25,3)` and focus is `x=25 1/12`

Explanation:

When the equation of a parabola is of the form `x=a(y-k)^2+h`, the vertex is `(h,k)`, axis of symmetry is `y-k=0` and focus is `(h+1/(4a),k)` and directrix is `x=h-1/(4a)`

As `x=-3y^2+12y+13` is a quadratic equation and can be converted into vertex form as follows

`x=-3y^2+12y+13`

= `-3(y^2-4y+4-4)+12`

= `-3(y-2)^2+12+13`

= `-3(y-2)^2+25`

Its vertex is `(25,3)` and axis of symmetry is `y=2`

Its focus is `(25+1/(4xx(-3)),2)` i.e. `(24 11/12,2)` and

diectrix is `x=25-1/(4xx(-3))=25 1/12`
graph{x=-3y^2+12y+13 [-39.83, 40.17, -20.16, 19.84]}