x^(2)-(y-6)^(2)=36 and y=-x^(2)
Solve the equation
You want find the intersection of the two curves
x^2 - (y-6)^2 = 36, which is a hyperbola, with
y = -x^2 , which is a downwards parabola
use substitution.
form y = -x^2, x^2 = -y
into the first:
-y - (y^2 - 12y + 36) = 36
-y - y^2 + 12y - 36 - 36 = 0
y^2 -11y + 72 = 0
There are no real solutions to this equation, so
there is no solution to the system of equations,
check:
https://www.wolframalpha.com/input/?i=plot+x%5E2-(y-6)%5E2%3D36,+y%3D-x%5E2
To solve the given equation, we can substitute the value of y from the second equation into the first equation. Let's substitute y with -x^2:
x^2 - (y-6)^2 = 36
x^2 - (-x^2-6)^2 = 36
x^2 - (x^4 + 12x^2 + 36) = 36
Now we can simplify the equation by expanding the square term:
x^2 - x^4 - 12x^2 - 36 = 36
-x^4 - 11x^2 - 36 = 0
This equation is a quadratic equation in terms of x^2. To solve it, let's substitute x^2 with a new variable, such as u:
-u^2 - 11u - 36 = 0
Now, we can solve this quadratic equation by factoring, completing the square, or using the quadratic formula. Let's use factoring for this example:
(-u - 9)(u + 4) = 0
Now we have two factors that can be set equal to zero:
-u - 9 = 0 --> u = -9
u + 4 = 0 --> u = -4
Since u represents x^2, we can solve for x:
When u = -9:
x^2 = -9
x = ±√(-9)
We have complex solutions here because the square root of a negative number is imaginary. Therefore, the solutions for this case are:
x1 = √9 * i = 3i
x2 = -√9 * i = -3i
When u = -4:
x^2 = -4
x = ±√(-4)
Similarly, we have complex solutions for this case as well:
x3 = 2i
x4 = -2i
Therefore, the solutions to the given equation are:
x1 = 3i
x2 = -3i
x3 = 2i
x4 = -2i