Points of Intersection of Parabola

[nycmath]

See attachment.

[Math Tutor]

Substitute x^2 = y from the parabola into the ellipse:

[tex]b^2y+a^2y^2=a^2b^2\quad\Longrightarrow\quad a^2y^2+b^2y-a^2b^2=0[/tex]

This is a quadratic in y:

[tex]y=\frac{-b^2\pm b\sqrt{b^2+4a^4}}{2a^2}[/tex]

Since y = x^2 must be non-negative, only the "+" root works:

[tex]y_0=\frac{b\left(\sqrt{b^2+4a^4}-b\right)}{2a^2},\qquad x=\pm\sqrt{y_0}[/tex]

So for any a, b > 0 the curves meet in exactly two points, symmetric about the y-axis:

[tex]\left(\pm\sqrt{\frac{b\left(\sqrt{b^2+4a^4}-b\right)}{2a^2}},\ \frac{b\left(\sqrt{b^2+4a^4}-b\right)}{2a^2}\right)[/tex]

Check with a = 2, b = 1 (ellipse x^2/4 + y^2 = 1): 4y^2 + y - 4 = 0, y_0 = (sqrt(65)-1)/8 = 0.8828, x = +/- 0.9396.

[nycmath]

Substitute x^2 = y from the parabola into the ellipse:

[tex]b^2y+a^2y^2=a^2b^2\quad\Longrightarrow\quad a^2y^2+b^2y-a^2b^2=0[/tex]

This is a quadratic in y:

[tex]y=\frac{-b^2\pm b\sqrt{b^2+4a^4}}{2a^2}[/tex]

Since y = x^2 must be non-negative, only the "+" root works:

[tex]y_0=\frac{b\left(\sqrt{b^2+4a^4}-b\right)}{2a^2},\qquad x=\pm\sqrt{y_0}[/tex]

So for any a, b > 0 the curves meet in exactly two points, symmetric about the y-axis:

[tex]\left(\pm\sqrt{\frac{b\left(\sqrt{b^2+4a^4}-b\right)}{2a^2}},\ \frac{b\left(\sqrt{b^2+4a^4}-b\right)}{2a^2}\right)[/tex]

Check with a = 2, b = 1 (ellipse x^2/4 + y^2 = 1): 4y^2 + y - 4 = 0, y_0 = (sqrt(65)-1)/8 = 0.8828, x = +/- 0.9396.

You outdid yourself. This is too perfect. Are you a math professor?