Question

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`?

Answer 1

`(x, y) = (-4, 0), (-4, 4), (6, 1) and (6, 3)`. A nice Socratic graph illustrates the crossings of the hyperbola and the ellipse represented by the given equations.

Explanation:

Add and subtract.

`x^2-2x-24=0`, giving `x -4 and 6`.

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 `x = 6, y^2-4y+3= 0`, giving the common points (6, 1) and (6, 3).

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]}