Impartial differentiation?

Equation of the curve:
(x/2)² + (y/2)² + yx² = 5
[(x/2)² + (y/2)² + yx²] * 4 = 5 * 4
x² + y² + 4yx² = 20

When y = 2:
x² + 2² + 4(2)x² = 20
x² + 4 + 8x² = 20
x² = 16/9
x = ±4/3

The points of contact are (4/3, 2) and (-4/3, 2).

Differentiate the equation of the curve on the both sides.
(d/dx) (x² + y² + 4yx²) = (d/dx) 20
2x + 2yy' + 8xy + 4x²y' = 0
2yy' + 4x²y' = -2x - 8xy
y' = (-2x - 8xy) / (2y + 4x²)
y' = -(x + 4xy) / (y + 2x²)

For the tangent at (4/3, 2):
Slope = -[(4/3) + 4(4/3)(2)]/[2 + 2(4/3)²] = -54/25
Equation:
y - 2 = (-54/25) [x - (4/3)]
y - 2 = -(54x/25) + (72/25)
y = -(54x/25) + (122/25) …… [1]

For the tangent at (-4/3, 2):
Slope = -[(-4/3) + 4(-4/3)(2)]/[2 + 2(-4/3)²] = 54/25
Equation:
y - 2 = (54/25) [x + (4/3)]
y - 2 = (54x/25) + (72/25)
y = (54x/25) + (122/25) …… [2]

[1] = [2]:
-(54x/25) + (122/25) = (54x/25) + (122/25)
x = 0

Substitute x = 0 into [1]:
y = 122/25

Coordinates of P = (0, 122/25)

when y = 2 then you should see that x = ± 4 / 3 and the slopes of the T-lines is
± 108/59 ......use these to find the T_line equations and then find the point P