How do you find the exact solutions to the system #3x^2-20y^2-12x+80y-96=0# and #3x^2+20y^2=80y+48#?
1 Answer
Dec 24, 2016
Explanation:
Add and subtract.
10y^2-40y+12+3x=0#, giving,
at #x = -4, y(y-4)=0 that gives common points (-4, 0) and (-4, 4) and ,
at
The given equations represent a hyperbola and an ellipse,
respectively
graph{(3x^2-20y^2-12x+80y-96)(3x^2+20y^2-80y-48)=02 [-10, 10, -5, 5]}