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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.

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    For this kind of problem you should draw a picture before you tackle the algebra. Then you will see the four points and the two $\pm$ signs. – Ethan Bolker May 01 '23 at 20:17

2 Answers2

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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.

SomeProgrammer
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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.

Heldert V
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