Question

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

Answer 1

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 `a = 1`, `h = 1` and `k=3`

This is the vertex form of the equation of a parabola with vertical axis, vertex `(h, k) = (1, 3)` and leading coefficient `1`.

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 `(h, k)` is the centre and `r` is the radius.

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` and `y` axes can be written in the form:

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

where `(h, k)` is the centre, `a` is the radius along the horizontal axis and `b` is the radius along the vertical axis.

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 `(h, k)` is the centre, `a` is the distance between the centre and the vertices and the slopes of the asymptotes are `+-b/a`.

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.