A curve passing through the point (1,1) has the property that the perpendicular distance of the origin from the normal at any point P of the curve is equal to the distance of p from the x-axis. Determine the equation of the curve.

Use y' to represent the derivative of the function y = f(x).
Then the equation of the normal at a point (t, f(t)) is
y - f(t) = (-1/y')(x - t) where y' = f '(t).
i.e.
yy' - y' f(t) = -x + t
yy' + x - y' f(t) - t = 0
Using the formula for the perpendicular distance of a point from a line, the distance of the origin from this normal is
|y' f(t) + t| / √(1 + (y')²)
We can now revert to using (x, y) for (t, f(t)) since we are no longer concerned with coordinates of points on the normal. The distance of P from the x axis is y [actualy |y|, but it doesn't matter since we're going to square it], and so the required equation is
y = |y y' + x| / √(1 + (y')²)
y² = (y y' + x)² / (1 + (y')²)
y² + y² (y')² = y² (y')² + 2xyy' + x²
y² = 2xyy' + x²
Substitute y = ux,
therefore y' = u + u' x,
and 2xyy' = 2ux²(u + u' x)
hence
u²x² = 2ux²(u + u' x) + x²
u² = 2u(u + u' x) + 1
.. = 2u² + 2uu'x + 1
2uu'x = -u² - 1
2u u' .. ... - 1
-------- .= .------
u² + 1 ...... x
[That's meant to be 2uu' / (u² + 1) = - 1/x]
Integrate with respect to x:
ln(u² + 1) = - ln x + C
substitute x = 1, u = 1/1 = 1:
C = ln 2
Therefore
ln (u² + 1) = ln 2 - ln x
Therefore
u² + 1 = 2/x
Multiply both sides by x², and use ux = y:
y² + x² = 2x
This is the circle with centre (1, 0), radius 1.
Checking back, we find that the normal to the circle at the point (x, y) is the diameter at that point, and its distance from the origin is equal to y.