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

Question

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.

Answer 1

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.