# How do you identify the following equation #y = (x-1)^2 + 3# as a circle, parabola, ellipse or hyperbola?

##### 1 Answer

Compare with the vertex form of the equation of a parabola to see that it is indeed a parabola.

#### Explanation:

#y = (x-1)^2+3#

This equation is in the form:

#y = a(x-h)^2 + k#

with

This is the vertex form of the equation of a parabola with vertical axis, vertex

graph{y = (x-1)^2+3 [-8.75, 11.25, -1.44, 8.56]}

For comparison, note that the equation of a circle can be written in the form:

#(x-h)^2+(y-k)^2=r^2#

where

graph{((x-3)^2+(y-2)^2-4^2)((x-3)^2+(y-2)^2-0.01) = 0 [-8.29, 11.71, -3.16, 6.84]}

An ellipse with major and minor axes parallel to the

#(x-h)^2/(a^2)+(y-k)^2/b^2 = 1#

where

graph{((x-3)^2/4^2+(y-2)^2/3^2-1)((x-3)^2+(y-2)^2-0.01) = 0 [-8.29, 11.71, -3.16, 6.84]}

Reverse the sign and it becomes a hyperbola with horizontal axis

#(x-h)^2/(a^2)-(y-k)^2/b^2 = 1#

where

graph{((x-3)^2/4^2-(y-2)^2/3^2-1)((x-3)^2+(y-2)^2-0.01)(3(x-3)-4(y-2))(3(x-3)+4(y-2)) = 0 [-8.29, 11.71, -3.16, 6.84]}

Note that the equation of a hyperbola is one of the most varied.