How many points of intersection are there between the graphs of the hyperbola and ellipse?

3x^2+4y^2+3x−2y+3=0

3x^2−2y^2+3x−y+3=0

Guest Jan 15, 2020

#1**+2 **

Equate them.....because when they are equal is when they have an intersection/common point

3x^2+4y^2+3x−2y+3=3x^2−2y^2+3x−y+3 delete 3x^2 and 3x from both sides

4y^2 -2y+3 = -2y^2-y+3 gather like terms on one side

6y^2 -y = 0

y(6y-1) = 0 y = 0 or 1/6 now that you have values for 'y' sub back into one of the original equations to fing the corresponding 'x' values.

SO there is 2 points of intersection.

ElectricPavlov Jan 15, 2020

#2**+1 **

This is an interesting problem because the first equation: 3x^{2} + 4y^{2} + 3x - 2y + 3 = 0 does not have a real-valued graph.

If you attempt to place this graphing form: 3x^{2} + 3x + 4y^{2 }- 2y + 3 = 0

3(x^{2} + x) + 4(y^{2} - ½y) = -3

Completing the squares: 3(x^{2} + x + ¼) + 4(y^{2} - ½y + 1/16) = -3 + 3/4 + 1/4

3(x + ½)^{2} + 4(y - ¼)^{2} = -2

This is not a real-valued function.

So, the points of intersection do not exist in a real-valued plane.

geno3141 Jan 15, 2020

#3**+1 **

Interesting..... I was going to graph them on desmos....just had not done so yet!

ElectricPavlov
Jan 15, 2020