Y = X2 forms a parabola
2
(x, y) = (-3, -3) or (3, 3)
89
The solutions are: x = 4, y = 2 and x = -4, y = -2
They touch each other at (0, 100) on the x and y axis.
2
(x, y) = (-3, -3) or (3, 3)
89
y = x - 4 x^(2) + y^(2) = 8 Hence y^(2) = 8 - x^(2) Square up the linear eq'n and substitute Hence y^(2) = ( x - 4)^(2) NB Notice the RHS is wholly squared , not individual terms. Substitute ; this eliminates 'y' ( x - 4)^(2) = 8 - x^(2) Square out the brackets x^(2) - 8x + 16 = 8 - x^(2) Collect terms 2x^(2) -8x +8 = 0 Factor 2(x^(2) - 4x + 4) = 0 (x- 2)(x - 2) = 0 Hence x = 2 ( The 'x' coordinate) . Substitute into either eq;n for 'y' y = 2 - 4 y = -2 Hence ( x,y) = ( 2,-2) ; The point of contact of the tangent and the curve.
The solutions are: x = 4, y = 2 and x = -4, y = -2
y equals x plus 4 when y equals 0 then x equals 2i i is the square root of negative 1
The two rational solutions are (0,0,0) and (1,1,1). There are no other real solutions.
They touch each other at (0, 100) on the x and y axis.
Y = 4/x^2
If x = sin θ and y = cos θ then: sin² θ + cos² θ = 1 → x² + y² = 1 → x² = 1 - y²
They are +/- 5*sqrt(2)
Two equations: a linear one in y and a quadratic in x.