A tutorial on finding the points of intersection of a circle and an ellipse given by their equations.
We need to solve the system of equations given above.
We first multiply all terms of the second equation by \( - 4 \) and simplify to obtain:
\( x^2 + y^2 = 4 \)
\( -x^2 - \left(\frac{4}{9}\right) \left(y - 1\right)^2 = -4 \)
We now add side by side the two equations above to obtain a linear equation
\( y^2 - \left(\frac{4}{9}\right) \left(y - 1\right)^2 = 0 \)
Which may be written as
\( 5y^2 + 8y - 4 = 0 \)
Solve the quadratic equation for y to obtain two solutions
\( y = -2 \) and \( \frac{2}{5} \)
We now substitute the values of \( y = -2 \) already obtained into the equation \( x^2 + y^2 = 4 \) and solve it for x as follows
\( x^2 + (-2)^2 = 4 \)
\( x = 0 \)
We now substitute the values of \( y = \frac{2}{5} \) already obtained into the equation \( x^2 + y^2 = 4 \) and solve it for x as follows
\( x^2 + \left(\frac{2}{5}\right)^2 = 4 \)
\( x = \frac{4\sqrt{6}}{5} \approx 1.96 \) and \( x = -\frac{4\sqrt{6}}{5} \approx -1.96 \)
The points of intersection of the ellipse and the circle are
\( (-2 , 0) \); \( \left(-\frac{4\sqrt{6}}{5} , \frac{2}{5}\right) \); \( \left(\frac{4\sqrt{6}}{5} , \frac{2}{5}\right) \)
Shown below is the graph of a circle and an ellipse and their points of intersection.
