THE LINE Y=X+4 INTERSECTS THE CURVE
2X^2 + 3XY - Y^2 + 1 = 0
To find the intersection points between the line y = x + 4 and the curve 2x^2 + 3xy - y^2 + 1 = 0, we need to solve the equations simultaneously.
Step 1: Substitute y = x + 4 into the equation of the curve:
2x^2 + 3x(x + 4) - (x + 4)^2 + 1 = 0
Step 2: Simplify the equation:
2x^2 + 3x^2 + 12x - (x^2 + 8x + 16) + 1 = 0
Combine like terms:
4x^2 + 4x - 15 = 0
Step 3: Solve the quadratic equation for x.
To solve the quadratic equation, you can use factoring, completing the square, or the quadratic formula. In this case, we will use the quadratic formula.
The quadratic formula is:
x = (-b ± √(b^2 - 4ac)) / 2a
For the equation 4x^2 + 4x - 15 = 0, the values of a, b, and c are:
a = 4, b = 4, c = -15
Substituting these values into the quadratic formula:
x = (-(4) ± √((4)^2 - 4(4)(-15))) / (2(4))
Simplifying:
x = (-4 ± √(16 + 240)) / 8
x = (-4 ± √256) / 8
x = (-4 ± 16) / 8
This gives us two possible values for x:
x1 = (-4 + 16) / 8 = 12/8 = 3/2
x2 = (-4 - 16) / 8 = -20/8 = -5/2
Step 4: Substitute the x-values back into the equation y = x + 4 to find the corresponding y-values.
For x1 = 3/2:
y1 = (3/2) + 4 = 11/2
For x2 = -5/2:
y2 = (-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).