# Maximum number of intersections of two quadratics

• Apr 6th 2010, 07:10 AM
r45
Maximum number of intersections of two quadratics
Hi there,

I am wondering what is the maximum number of points of intersection of 2 quadratics of the form:

$ax^2 + bxy + cy^2 + dx + ey + f = 0$

I think the answer is 4, because expressing y in terms of x has a $\pm$ sign, and then subbing each of these y-values in to the other quadratic leads to 2 solutions each through use of the quadratic formula. Thus 2 x 2 = 4 is the maximum number of intersections.

This also seems intuitive, if we imagine for example a hyperbola being intersected 4 times by an ellipse. I can't picture in my head any 2 quadratics who intersect more than 4 times.

Anyway if someone could please verify or disprove my assertion I'd be very grateful!
• Apr 6th 2010, 07:25 AM
Amer
Quote:

Originally Posted by r45
Hi there,

I am wondering what is the maximum number of points of intersection of 2 quadratics of the form:

$ax^2 + bxy + cy^2 + dx + ey + f = 0$

I think the answer is 4, because expressing y in terms of x has a $\pm$ sign, and then subbing each of these y-values in to the other quadratic leads to 2 solutions each through use of the quadratic formula. Thus 2 x 2 = 4 is the maximum number of intersections.

This also seems intuitive, if we imagine for example a hyperbola being intersected 4 times by an ellipse. I can't picture in my head any 2 quadratics who intersect more than 4 times.

Anyway if someone could please verify or disprove my assertion I'd be very grateful!

infinite number of points take the same quadratic for the two curves
if the two curves are different you have 4 points

$ax^2 + bxy + cy^2 + dx + ey + f = 0$

if we take two curves say

$a_1x^2 +b_1xy + c_1 y^2 + d_1 x + e_1 y + f_1 = 0$...(1)

$a_2x^2 +b_2xy + c_2 y^2 + d_2 x + e_2 y + f_2 = 0$ ...(2)

rewrite the first one like

$c_1 y^2+ e_1 y +b_1xy + d_1 x +a_1x^2 + f_1 = 0$

$c_1 y^2 + y(e_1 + b_1x) + d_1x + a_1x^2 +f_1 = 0$

solve for y

$y = \frac{-(e_1 + b_1x) \pm \sqrt{(e_1 + b_1x)^2 - 4c_1( d_1x + a_1x^2 +f_1)}}{2c_1}$

now we have to curves from (1)

$y_1 = \frac{-(e_1 + b_1x) + \sqrt{(e_1 + b_1x)^2 - 4c_1( d_1x + a_1x^2 +f_1)}}{2c_1}$

$y_2 = \frac{-(e_1 + b_1x) - \sqrt{(e_1 + b_1x)^2 - 4c_1( d_1x + a_1x^2 +f_1)}}{2c_1}$

and these two curve dose not intersect with each other you can see that

do the same thing for (2) you will have two curves

$y_3 =\frac{-(e_2 + b_2x) + \sqrt{(e_1 + b_1x)^2 - 4c_2( d_2x + a_2x^2 +f_2)}}{2c_2}$

$y_4 = \frac{-(e_2 + b_2x) - \sqrt{(e_2 + b_2x)^2 - 4c_2( d_2x + a_2x^2 +f_2)}}{2c_2}$

and these two never intersect