from the hyperbola: x^2 = y^2 + 4
sub into the ellipse
2(y^2+4) + 3y^2 = 48
5y^2 = 40
y^2 = 8
y = ±√8 or ± 2√2
the x^2 = y^2 + 4
= 8+4 = 12
x = ±√12 or ± 2√3 ---> your 4 points
enter as instructed.
http://www.wolframalpha.com/input/?i=2x%5E2%2B3y%5E2%3D48+%2C+x%5E2-y%5E2%3D4
Neat, eh ?
order your answers from smallest to largest x, then from smallest to largest y. If there is no solution, enter NO Solution
2x^2+3y^2=48
x^2-y^2=4
(x,y)= ( , ) (, )
(x,y)= ( , ) (, )
(x,y)= ( , ) (, )
(x,y)= ( , ) (, )
1 answer