THE LINE Y=X+4 INTERSECTS THE CURVE
2X^2 + 3XY - Y^2 + 1 = 0
So it does.
use substitution to solve ...
2x^2 + 3x(x+4) - (x+4)^2 + 1 = 0
2x^2 + 3x^2 + 12x - x^2 - 8x - 16 + 1 = 0
4x^2 + 4x - 15 = 0
(2x-3)(2x+5) = 0
I am sure you can finish it.
To find the points of intersection between the line y = x + 4 and the curve 2x^2 + 3xy - y^2 + 1 = 0, we can substitute the equation of the line into the equation of the curve.
First, substitute y = x + 4 into the equation 2x^2 + 3xy - y^2 + 1 = 0:
2x^2 + 3x(x + 4) - (x + 4)^2 + 1 = 0
Simplifying this equation will give us a quadratic equation. Let's expand and collect like terms:
2x^2 + 3x^2 + 12x - x^2 - 8x - 16 + 1 = 0
4x^2 + 4x - 15 = 0
Now we have a quadratic equation. We can solve this equation by factoring, completing the square, or by using the quadratic formula.
Since factoring may not be straightforward in this case, let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
For our equation 4x^2 + 4x - 15 = 0, a = 4, b = 4, and c = -15.
Substituting these values into the quadratic formula:
x = (-4 ± √(4^2 - 4*4*(-15))) / (2*4)
x = (-4 ± √(16 + 240)) / 8
x = (-4 ± √256) / 8
x = (-4 ± 16) / 8
Now we have two possible values for x:
x₁ = (-4 + 16) / 8 = 12 / 8 = 3/2
x₂ = (-4 - 16) / 8 = -20 / 8 = -5/2
Now substitute these x-values back into the equation y = x + 4 to find the corresponding y-values:
For x = 3/2:
y = 3/2 + 4 = 11/2
For x = -5/2:
y = -5/2 + 4 = 3/2
Therefore, the line y = x + 4 intersects the curve 2x^2 + 3xy - y^2 + 1 = 0 at the points (3/2, 11/2) and (-5/2, 3/2).