Unlike the previous activities, I know have two conics. In general, intersections of two conics can be quite difficult: four intersection points are possible, and sometimes the solution require dealing with a degree 4 polynomial. Where possible, I’ll look for algebraic or geometric tricks to make the process easier.
One important algebraic trick is substituting an expression, instead of just the variable. In the two equations for this activity, \(x\) shows up only as \(x^2\text{,}\) and likewise \(y\) shows up only as \(y^2\text{.}\) Therefore, I can isolate \(x^2\) (or, if I wanted, \(y^2\)) in one equation and substitute that in the other equation. I’ll use the circle equation first.
\begin{equation*}
x^2 + y^2 = 16 \implies x^2 = 16 - y^2
\end{equation*}
Then I can use this expression to replace the \(x^2\) in the equation of the ellipse. That will give a quadratic in \(y\text{,}\) which I can then solve.
\begin{align*}
\frac{(16-y^2)}{16} + \frac{y^2}{4} \amp = 1 \\
1 - \frac{y^2}{16} + \frac{y^2}{4} \amp = 1\\
\frac{3y^2}{16} = 0 \implies y = 0
\end{align*}
With \(y=0\text{,}\) substituting this value for \(y\) in either equation will result in \(x = \pm 4\text{.}\) Therefore, there are two intersection points: \((4,0)\) and \((-4,0)\text{.}\)
There is an alternative approach here as well. It is often good to think about the geometry when solving intersection problems. In this problem, I have two conics centred at the origin. There is a circle with radius \(4\) and an ellipse with major axis of \(4\) in the \(x\) direction and minor axis of \(2\) in the \(y\) direction. Therefore, from the geometry alone, I can conclude that the ellipse sits inside the circle but that they both reach out to the points \((4,0)\) and \((-4,0)\) along the \(x\) axis. This gives me a geometric argument for the intersection points without any calculation at all.
