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

Jan 18, 2016

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

#### Explanation:

$y = {\left(x - 1\right)}^{2} + 3$

This equation is in the form:

$y = a {\left(x - h\right)}^{2} + k$

with $a = 1$, $h = 1$ and $k = 3$

This is the vertex form of the equation of a parabola with vertical axis, vertex $\left(h , k\right) = \left(1 , 3\right)$ 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:

${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2}$

where $\left(h , k\right)$ 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:

${\left(x - h\right)}^{2} / \left({a}^{2}\right) + {\left(y - k\right)}^{2} / {b}^{2} = 1$

where $\left(h , k\right)$ 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

${\left(x - h\right)}^{2} / \left({a}^{2}\right) - {\left(y - k\right)}^{2} / {b}^{2} = 1$

where $\left(h , k\right)$ is the centre, $a$ is the distance between the centre and the vertices and the slopes of the asymptotes are $\pm \frac{b}{a}$.

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.