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Consider the system $$y=x^2$$ and $$x^2 + y^2 = a $$for $x>0$, $y>0$, $a>0$.

Solving for equations give me $y+y^2 = a$, and ultimately $$y = \frac {-1 + \sqrt {4a+1}} {2} $$ (rejected $\frac {-1 - \sqrt {4a+1}} {2} $ since $y>0$).

The next part is to plot on the $x-y$ plane for different values of $a$. Is plotting the graph of $y = x^2$ insufficient?

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First plot the parabola $y=x^{2}$, then plot the semicircle $y = sqrt(a-x^{2})$ which lies above $x$ axis. You'll get $2$ points of intersection. – Hyperbola Jan 31 '13 at 4:15
@Hyperbola $x>0$, so only one point – apnorton Jan 31 '13 at 4:17
Sorry for that. – Hyperbola Jan 31 '13 at 4:18
up vote 1 down vote accepted

Yes, it is insufficient.

You should notice that this equation is "special:" $$x^2 + y^2 = a$$ This is the graph of a circle, radius $\sqrt{a}$.

So, your graph should contain both the parabola and the part of the circle in the region in question.

Here's a link to a graph from Wolfram Alpha which may help give some intuition. The darkest shaded region that is there is the region of interest.

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