Calculus of a Single Variable, Chapter 6, 6.3, Section 6.3, Problem 25

We need to find the equation of the graph that passes through the point (0,2). The slope of the tangent line to the graph of y(x) is the derivative y' and it is given by the equation
y' = x/(4y) .
This equation can be solved by the method of separating variables. First, rewrite
(dy)/(dx) = x/(4y) .
Now, multiply by 4y and dx:
4ydy = xdx
Now both sides can be integrated:
4y^2/2 = x^2/2 + C . Here, C is an arbitrary constant.
From here,
4y^2 = x^2 + 2C
Since the graph of the equation has to pass through the point (0, 2), we can find C:
4*2^2 = 0 + 2C
C = 8.
So the equation of the graph is
4y^2 = x^2 + 16 , which can also be written as
y^2/4 - x^2/16 = 1 . This is a hyperbola that opens up and down, with the vertices at the points (0, 2) and (0, -2).
y^2/4 = x^2/16 + 1
y^2 = x^2/4 + 4
y = +-sqrt(x^2/4 + 4)

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