Your question is quite general. It does not lend itself to a clear concise answer. However, a few general features of the solutions can be outlined.
For the case of a parabola and and ellipse, there can be: 1) No points of intersection, 2) a single point of intersection (this happens when the parabola is tangent to the ellipse, 3) two points of intersection. A solution to a specific problem can be obtained by the method of substitution followed by solution of a quadratic equation. The equation for the parabola will be something like y = 3 x^{2} + 2x + 7. To get a solution, one substitutes for y in the equation for the ellipse using the parabola equation. The resulting equation will be a quadratic equation involving only x. This can be solved for x using the quadratic formula. One will get zero, one or two solutions for x, which can be substituted back into the parabola equation to get corresponding y values. The (x,y) pairs are the points of intersection.
For the case of two ellipses, there can be: 1) No points of intersection, 2) one point of intersection (this can happen when one ellipse is outside the other and tangent to it, 3) two points of intersection (this can happen two ways - first: the two ellipses overlap near there edges - second: one ellipse is inside the other and tangent two it at two points), 3) three points of intersection (one of the three is a point of tangents ), 5) four points of intersection ( the ellipses overlap ). Not all of these are possible if both ellipses are centered at the same origin. The method of substitution can be used to get solutions, but can be a lot of work for the general case. If both ellipses are centered at the origin, one can solve one equation for y^{2} and substitute into the other to get an equation quadratic in x. This can be solved by using the quadratic formula. The results for x ( none, one or two), can be substituted back into the first equation to yield a simple quadratic equation for y. The result for y (none, one or two) will yield up to two coordinate points of intersection for each value of x. In this way one can get up to four points of intersection.
Alternatively, graphing calculators can be used to solve these problems. The graphs will show which of the cases discussed above is the case for any particular problem.
Comments