# A function with the same slope as $b\sqrt{\frac{x^2}{a^2}-1}$ but not imaginary in [0,a]?

For some fixed $a,b \in \mathbb{R}$, $y = b\sqrt{\frac{x^2}{a^2}-1}$ is supposed to plot the boundary of an ellipse in $\left[0,a\right]$. I came up with that function but it has the defect that it runs into imaginary numbers for $x < a$. I would like to calculate the area of the ellipsis by quadrupling the integral in $\left[0,a\right]$. Would you mind giving me some advice on how to deal with the imaginary part? Maybe there's some way of avoiding it altogether (by using a slightly different function).

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$y = b\sqrt{\frac{x^2}{a^2}-1}$ is a hyperbola. You probably want to plot $y = b\sqrt{1-\frac{x^2}{a^2}}$. – Sidharth Iyer May 12 '12 at 9:40