Find the points where these quadratic relations intersect. 16x^2-5y^2=64, 16x^2+25y^2-96x=256
just substitute:
16x^2 - 96x + 5(16x^2-64) = 256
16x^2 - 96x + 80x^2 - 320 = 256
96x^2 - 96x - 576 = 0
x^2 - x - 6 = 0
(x+2)(x-3) = 0
Now you have x, get y.
see the solution at
http://www.wolframalpha.com/input/?i=solve+16x^2-5y^2%3D64%2C+16x^2%2B25y^2-96x%3D256
subtract them :
30y^2 - 96x = 192
5y^2 = 32 + 16x
back into the first:
16x^2 - (32+16x) = 64
16x^2 -16x - 96 = 0
x^2 - x - 6 = 0
(x-3)(x+2) = 0
x = 3 or x = -2
if x = 3,
5y^2 = 32+48 = 80
y^2 = 16
y = ± 4
if x = -2
5y^2 = 32-32 = 0
y = 0
so we have 3 points:
(-2,0) , (3,4) and (3,-4)
verification:
http://www.wolframalpha.com/input/?i=solve++16x%5E2-5y%5E2%3D64%2C+16x%5E2%2B25y%5E2-96x%3D256
To find the points where these quadratic relations intersect, we can solve the given equations simultaneously.
The two equations are:
1) 16x^2 - 5y^2 = 64
2) 16x^2 + 25y^2 - 96x = 256
Let's solve them step by step using the method of substitution.
Step 1: Simplify Equation 2
Rearrange Equation 2 to isolate the variable "x":
16x^2 + 25y^2 - 96x = 256
16x^2 - 96x + 25y^2 = 256
16(x^2 - 6x) + 25y^2 = 256
16(x^2 - 6x + 9) + 25y^2 = 256 + 16(9)
16(x - 3)^2 + 25y^2 = 400
Divide by 400 to make the right side equal to 1:
[(x - 3)^2]/25 + [y^2]/(400/25) = 1
[(x - 3)^2]/(5^2) + [(y^2)/(20^2)] = 1
Now we have Equation 1 and Equation 2 in a standard form for ellipse equations.
Step 2: Comparing and Analyzing Equations
Equation 1 represents a hyperbola, while Equation 2 represents an ellipse. Both equations can be written in the standard form of an ellipse: (x-h)^2/a^2 + (y-k)^2/b^2 = 1
Comparing the equations, we can conclude that the center of the ellipse is at the point (h, k) = (3, 0), and the semi-major axis is a = 5, while the semi-minor axis is b = 20.
Step 3: Finding the Intersections
To find the points of intersection, we need to solve the system of equations formed by the hyperbola and the ellipse simultaneously.
Let's substitute the value of y^2 from Equation 1 into Equation 2:
16x^2 + 25(64 + 5y^2)/5 - 96x = 256
16x^2 + 64 + 5y^2 - 96x = 256
16x^2 - 96x + 5y^2 = 192
Now we have a system of equations:
16x^2 - 5y^2 = 64
16x^2 - 96x + 5y^2 = 192
Combining the like terms, we get:
32x^2 - 96x = 128
Dividing both sides by 32 to simplify:
x^2 - 3x = 4
Rearranging the equation:
x^2 - 3x - 4 = 0
Factoring the quadratic equation, we have:
(x - 4)(x + 1) = 0
So, we have two possible x-values: x = 4 and x = -1.
Substituting these x-values into Equation 1, we can solve for the corresponding y-values:
For x = 4:
16(4)^2 - 5y^2 = 64
16(16) - 5y^2 = 64
256 - 5y^2 = 64
-5y^2 = -192
y^2 = 38.4
y = ±√38.4
For x = -1:
16(-1)^2 - 5y^2 = 64
16 - 5y^2 = 64
-5y^2 = 48
y^2 = -9.6 (This is not a real value)
Therefore, the points where the quadratic relations intersect are approximately (4, √38.4) and (4, -√38.4).