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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

 Jan 15, 2020
 #1
avatar+19815 
+1

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.

 
 Jan 15, 2020
edited by ElectricPavlov  Jan 15, 2020
 #2
avatar+17893 
+1

This is an interesting problem because the first equation:  3x2 + 4y2 + 3x - 2y + 3 =  0  does not have a real-valued graph.

If you attempt to place this graphing form:      3x2 + 3x  + 4y- 2y + 3 =  0 

                                                                        3(x2 + x)  +  4(y2 - ½y)  =   -3

Completing the squares:                                 3(x2 + x + ¼)  +  4(y2 - 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.

 
 Jan 15, 2020
 #3
avatar+19815 
0

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

 
ElectricPavlov  Jan 15, 2020

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