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Alexandra [31]
3 years ago
14

What is the equation of the quadratic graph with a focus of (3, 1) and a directrix of y = 5?

Mathematics
1 answer:
telo118 [61]3 years ago
4 0

Answer:

f(x)=-\frac{1}{8}(x-3)^2+3

Step-by-step explanation:

We want to find the equation of the parabola with a focus of (3,1) and directrix y=5.

Considering the directrix, the quadratic graph must open downwards.


The equation of this parabola is given by the formula,

(x-h)^2=4p(y-k), where (h,k) is the vertex of the parabola.


The axis of this parabola meets the directrix at (3,5).

Since the vertex is the midpoint of the focus and the point of intersection of the axis of the parabola and the directrix,


h=\frac{3+3}{2}=3 and k=\frac{5+1}{2}=3.


The equation of the parabola now becomes,


(x-3)^2=4p(y-3).


Also |p| is the distance between the vertex and the directrix.


|p|=2


This implies that p=-2\:or\:2.


Since the parabola turns downwards,


p=-2.



Our equation now becomes,


(x-3)^2=4(-2)(y-3).


(x-3)^2=-8(y-3).


We make y the subject to get,

y=-\frac{1}{8}(x-3)^2+3).


This is the same as

f(x)=-\frac{1}{8}(x-3)^2+3).









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