10x^2+5 y^2-2xy-28x-6y+41=0 and 3x^2-2y^2+5xy-17x-6y+20
Take a look at wiki pedia for a good article on rotation of axes.
In this case, the discriminant B^2 - $AC < 0, so we have a rotated ellipse.
The angle t is such that tan(2t) = B/(A-C)
Letting c == cos(t) and s =sin(t) the new coordinates in x' and y' can be found by letting
x' = xx + ys
y' = -sx + yc
So, with A,B,C,D,E,F = 10 -2 5 -28 -6 41, we have a new ellipse (if my math is right)
(x-1.2931)2 / 2.19262 + (y-1.1635)2/3.19262 = 1
Typos: B^2 - 4AC
x' = xc + ys
y' = -xs + yc
To find the common solution to the system of equations:
1. 10x^2 + 5y^2 - 2xy - 28x - 6y + 41 = 0
2. 3x^2 - 2y^2 + 5xy - 17x - 6y + 20 = 0
We can use the method of substitution or elimination. I'll explain how to approach it using substitution:
Step 1: Solve one equation for one variable in terms of the other.
Let's solve equation 1 for x:
Rearrange equation 1 by isolating 'x':
10x^2 - 2xy - 28x = -5y^2 + 6y - 41
Now, factor out 'x':
x(10x - 2y - 28) = -5y^2 + 6y - 41
Divide throughout by (10x - 2y - 28):
x = (-5y^2 + 6y - 41) / (10x - 2y - 28) [Equation A]
Step 2: Substitute equation A into equation 2.
Replace 'x' in equation 2 with the expression we obtained from step 1:
3((-5y^2 + 6y - 41) / (10x - 2y - 28))^2 - 2y^2 + 5((-5y^2 + 6y - 41) / (10x - 2y - 28))y - 17((-5y^2 + 6y - 41) / (10x - 2y - 28)) - 6y + 20 = 0
After substituting and simplifying, we will have one equation in terms of 'y' only.
Step 3: Solve the equation obtained from step 2 for 'y'.
Simplify the equation and solve for 'y' using methods like factoring or the quadratic formula. The resulting equation might be a quadratic or linear equation in 'y'.
Step 4: Substitute the value of 'y' obtained from step 3 into equation A.
Once you have found the value(s) of 'y', substitute it back into equation A to determine the corresponding value(s) of 'x'.
These values of 'x' and 'y' will be the common solutions to the given system of equations.