Determine the point on the hyperbola −3x2+2y2=10 closest to the point (4, 0).

-3x^2 + 2y^2 = 10
2y^2 = 10 + 3x^2
y^2 = 5 + 3/2x^2

distance between two points is given by
d = sqrt((x2 - x1)^2 + (y2 - y1)^2) = sqrt((x - 4)^2 + (y - 0)^2) = sqrt((x - 4)^2 + y^2) = sqrt((x - 4)^2 + 5 + 3/2x^2)
For d to be minimum (closest), dd/dx = 0
(2(x - 4) + 3x)/2sqrt((x - 4)^2 + 5 + 3/2x^2) = 0
2(x - 4) + 3x = 0
2x - 8 + 3x = 0
5x = 8
x = 8/5 = 1.6

y^2 = 5 + 3/2(1.6)^2 = 5 + 3/2(2.56) = 221/25
y = sqrt(221/25) = 2.97

Therefore, the required point is (1.6, 2.97)

31 is prime. and it’s prime factorization is just 31.