Find all points of intersection of the ellipses?

Question

Find all points of intersection of the ellipses x^2/a^2+y^2/b^2=1 and x^2/b^2+y^2/a^2=1 where a>b.

this is a pre-calc question so no advance calc please heh

Answer 1

I think you can see quite easily that all the points of intersection in this particular case lie on a square with centre at (0,0), by symmetry.

So, if x = y = p at the first quadrant intersection, we have

b^2*p^2+ a^2*p^2 =a^2*b^2 and
a^2*p^2 + b^2*p^2 = a^2*b^2

Hence p^2 = a^2*b^2/[a^2 + b^2].

The points of intersection are (p,p), (-p,-p), (p,-p) and (-p,p).

Answer 2

x²/a² + y²/b² = 1
x²/b² + y²/a² = 1

Set the two equations equal.

x²/a² + y²/b² = x²/b² + y²/a²

Multiply thru by a²b² to clear the denominators.

b²x² + a²y² = a²x² + b²y²
a²y² - b²y² = a²x² - b²x²
(a² - b²)y² = (a² - b²)x²
y² = x²

y = ±x

Plug back into the first equation and solve for x.

x²/a² + y²/b² = 1
x²/a² + x²/b² = 1
x²(1/a² + 1/b²) = 1
x² = 1/(1/a² + 1/b²)
x² = a²b²/(a² + b²)

x = ±ab/√(a² + b²)
y = ±ab/√(a² + b²)

So the points of intersection are four.

(x,y) = [±ab/√(a² + b²), ±ab/√(a² + b²)]