# In an equation that looks like the standard form of an ellipse, what must the constant on the RHS equal for exactly one solution?

I am working on a homework question: What must be the value(s) of $c$ for the following equation to have exactly 1 solution?

The equation is of the standard form of the equation for an ellipse,

$\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=c$

First I thought that there is no $c$ for which there is only one solution since an ellipse has infinitely many solutions. However, now I think if $c=0$ then there might only be one solution. However, I don't know how to prove this. Also, what if $c <0$? Will it still be an ellipse? Thanks.

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What "c" are you talking about, anyway?? –  DonAntonio Feb 5 '13 at 3:44
Oops sorry, I wrote k instead of c in the equation. Fixed now. –  Chloe Gonzales Feb 5 '13 at 3:46
A sum of squares is always non-negative. So there is $1$ solution precisely if $c=0$. That $1$ point set can I guess be thought of as a degenerate ellipse. For $c\lt 0$ there cannot be an $x,y$ that work. –  André Nicolas Feb 5 '13 at 3:50
Thanks. You could have posted this as an answer :) –  Chloe Gonzales Feb 5 '13 at 3:57

By "solutions" do you mean points $\,(x,y)\,$ in the plane that satisfy the given equation? If you're working over the reals then obviously there are infinite points that satisfy the equation whenever $\,c>0\,$ , whereas there's only one when $\,c=0\,$ , as a sum of squares equals zero in the reals iff each of the summands is zero. Finally, for $\,c<0\,$ there are no solutions.
Yes I do mean points $(x,y)$ that satisfy the equation (real numbers).I understand your first two points, but can you elaborate why there will be no solutions if $c<0$? Thanks. –  Chloe Gonzales Feb 5 '13 at 3:55