How many points these ellipses intersect.

Question

If $a \neq b$, how many points theses ellipses $\frac{x²}{a²} + \frac{y²}{b²} = 1$, $\frac{x²}{b²} + \frac{y²}{a²} = 1$ intersect themselves.

My attempt: I did $y² = b² -\frac{x²}{a²}b² = a² -\frac{x²}{b²}a² $ and I found $x = +- \frac{ab}{\sqrt{a²+ b²}}$ But I'm confuse about what I do with this.

Answer: 4 points.

Answer 1

You have found 2 x-coordinates such that the corresponding y coordinates coincide, when you take the root to get said y-coordinate you have to account the $\pm$ again, therefore getting 2 possible y-coordinates; taking into account each posibility for x and y yield the 4 intersection points.

Answer 2

Your reasoning is correct.

Following the same line of reasoning you obtain:

$y = ± \frac{ab}{√(a^2+b^2)}$

Hence the points of intersections are:

$(\frac{ab}{√(a^2+b^2)}, \frac{ab}{√(a^2+b^2)})$

$(\frac{ab}{√(a^2+b^2)}, -\frac{ab}{√(a^2+b^2)})$

$(-\frac{ab}{√(a^2+b^2)}, \frac{ab}{√(a^2+b^2)})$

$(-\frac{ab}{√(a^2+b^2)}, -\frac{ab}{√(a^2+b^2)})$

This makes four points in total.