What is the equation in standard form of the parabola with a focus at (1,5) and a directrix of y= 7?

1 Answer
Feb 16, 2016

#y=-1/4*x^2+1/2*x+23/6#

Explanation:

Focus is at (1,5) and directrix is y=7. So the distance between focus and directrix is #7-5=2 units# Vertex is at the mid point between Focus and Directrix. So vertex co-ordinate is (1,6) . The parabola opens down as focus is below the Vertex. We know the equation of parabola is #y=a*(x-h)^2+k# where (h,k) is the vertex. Thus the Equation becomes #y=a*(x-1)^2+6# now #a=1/4*c#where c is the distance between vertex and directrix; which is here equal to 1 so #a= -1/4*1 =-1/4# (negative sign is as the parabola opens down) So the equation becomes #y=-1/4*(x-1)^2+6 or y=-1/4*x^2+1/2*x+23/6#graph{-1/4x^2+1/2x+23/6 [-10, 10, -5, 5]} [ans]