Find the real solutions
8x^2 + y^2 = 25
8X^2 - y^2 = 39
Is this no solution because I tried solving it and i got the square root of 32 for y, but when i plugged it in to get x there was a negative under the square root sign which means its no solution, right? please help...thanks!
The first one is an ellipse
The second one is a hyperbola
What do you mean by solutions?
Do you want values of x where y = 0 or what?
Oh, I see from your earlier questions, solve the intersections of ellipse and hyperbola.
Well both are centered on the origin, so the hyperbola either cuts the ellipse or misses it entirely.
okay thanks! so that means...there is no solution for solving it algebraically?
The ellipse is from x = -2 to x = +2 and y = -1 to y = + 1
The hyperbola is above y = 5 and below y = -5
THEREFORE THEY NEVER HIT
Let's say from the ellipse y^2 = (25 - 8 x^2)
then from the hyperbola
8 x^2 - (25 - 8 x^2) = 39
16 x^2 -25 = 39
16 x^2 = 64
x^2 = 4
x = 2, x = -2
y^2 = 25 - 8(4) = 25 - 32 = imaginary y
This is at the very ends of the ellipse, but the hyperbola does not go there
whoa, too quick
the ellipse is from x = -5/(2 sqrt 2) to +5/(2 sqrt 2) which is about -1.77 to +1.77 and y from -5 to +5
The hyperbola is left of - 2.2 and right of +2.2
They do not hit no matter how you look at them :)