Find integer coordinates that lie on the curve.
Solve for y
5y^2 + 2xy + 2x^2 - 20 = 0
y = (1/10)*(-2x ± √(4x^2 - 40x^2 + 400))
y = (1/10)*(-2x ± √(400 - 36x^2))
y = (1/5)*(-x ± √(100 - 9x^2))
-10/3 < x < 10/3
Possible x values: (-3,-2,-1,0,1,2,3)
Take the derivative wrt x of 2x^2 + 2xy + 5y^2 = 20:
4x + 2(y + x*y') + 10*y*y' = 0
4x + 2y + 2xy' + 10yy' = 0
xy' + 5yy' = -(2x + y)
y' = -(2x + y) / (x+5y)
y' is zero when y = -2x, plug into equation
2x^2 + 2x(-2x) + 5(-2x)^2 = 20
2x^2 - 4x^2 + 20x^4 = 20
10x^4 - x^2 - 10 = 0
x^2 = (1 ± √(401))/20
ignoring the minus option, x^2 is close to 21/20
Maximum y is when x = approx ± 1.025 or y is ± 2.11
Possible y values: (-2,-1,0,1,2)
With a limited number of (x,y) values to check, the only ones on the curve are:
(0,2), (-2,2), (0,-2), (2,-2)
The perimeter is 2*(2 + 2√5) = 4(1 + √5).
Perimeter and Desmos drawing
