Intersection of two ellipses with steps please

Question

(x-3)^2/49 + (y-4)^2/13 = 1 

 

and 

 

(x+10)^2/50 + (y-4)^2/10 = 1

Answer 1

to find the intersections, we simply solve the pair of equations for x and y.  Let's use elimination:

(x-3)^2 / 49 + (y-4)^2 / 13 = 1    <<eqn1>>

(x+10)^2 / 50 + (y-4)^2 / 10 = 1   <<eqn2>>

 

Multiply eqn2 by 10/13:  (this will allow us to eliminate the y term)

(x+10)^2 (1/65) + (y-4)^2 / 10 = 10/13

Subtract eqn1:

-(x-3)^2 / 49 - (y-4)^2 / 13 = -1

gives:

(x+10)^2 (1/65) - (x-3)^2 / 49 = 10/13 - 1

 

expand and simplify:

(x^2 + 20x + 100 ) / 65 - (x^2 - 6x + 9) / 49 = -3/13

0.0154x^2 + 0.308x + 1.538 - 0.0204x^2 + 0.122x - 0.184 = 0.231

 

0.005x^2 + 0.430x + 1.12 = 0

x^2 + 86 + 224 = 0

 

x = -2.69, -83.3

 

At this point, check my arithmetic for mistakes, then sub into either of the equations to get the y values.