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)= ( , ) (, )
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 ?
To solve the system of equations,
1. Start with the equation x^2 - y^2 = 4.
Rearrange it to get x^2 = y^2 + 4.
2. Substitute this expression for x^2 into the first equation, 2x^2 + 3y^2 = 48.
Replace x^2 with y^2 + 4.
The equation becomes 2(y^2 + 4) + 3y^2 = 48.
Simplify it to 2y^2 + 8 + 3y^2 = 48.
Combine like terms to get 5y^2 + 8 = 48.
Rearrange it to get 5y^2 = 40.
Divide both sides by 5 to isolate y^2: y^2 = 8.
Take the square root of both sides to solve for y: y = ±√8 = ±2√2.
So we have two possible values for y: y1 = 2√2 and y2 = -2√2.
3. Plug these values for y into the equation x^2 = y^2 + 4.
For y1 = 2√2, x^2 = (2√2)^2 + 4, x^2 = 8 + 4, x^2 = 12.
Take the square root of both sides to solve for x: x = ±√12 = ±2√3.
So when y = 2√2, we have two possible values for x: x1 = 2√3 and x2 = -2√3.
For y2 = -2√2, x^2 = (-2√2)^2 + 4, x^2 = 8 + 4, x^2 = 12.
Take the square root of both sides to solve for x: x = ±√12 = ±2√3.
So when y = -2√2, we have two possible values for x: x3 = 2√3 and x4 = -2√3.
4. Putting it all together, we have four solutions (x, y):
(x1, y1) = (2√3, 2√2)
(x2, y1) = (-2√3, 2√2)
(x3, y2) = (2√3, -2√2)
(x4, y2) = (-2√3, -2√2)
Now let's order the solutions:
Smallest x to Largest x:
x1 = 2√3
x2 = -2√3
x3 = 2√3
x4 = -2√3
Smallest y to Largest y:
y2 = -2√2
y1 = 2√2
y1 = 2√2
y2 = -2√2
Therefore, ordered from smallest to largest x and y, the solutions are:
(2√3, -2√2), (-2√3, 2√2), (2√3, 2√2), (-2√3, -2√2)