How do I graph the ellipse with the equation $(x−4)^2/36+(y-3)^2/36=1$ ?

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Answer 1 · 2015-02-06T21:13:57

First of all... this is not an ellipse, o better... this is a very particular ellipse. This is a circle!

An ellipse have an equation like this:

$(x-x_c)^2/a^2+ (y-y_c)^2/b^2=1$ , where $C(x_c,y_c)$ is the center, $a$ is the orizontal semi-axis and $b$ is the vertical semi-axis.

But... if $a=b$ , the semi-axes are equal and so it is a circle.

In our case:

$(x-4)^2/36+(y-3)^2/36=1rArr(x-4)^2+(y-3)^2=36$ is the equation of a circle with centre $C(4,3)$ and radius $6$ , and thisis its graph:

graph{(x-4)^2+(y-3)^2=36 [-20, 20, -10, 10]}

How do I graph the ellipse with the equation (x−4)^2/36+(y-3)^2/36=1(x−4)^2/36+(y-3)^2/36=1?