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

1 Answer
Feb 12, 2017

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]}