Posted by rolan on Friday, August 10, 2012 at 5:46am.
Let AB be the base, which will be a fixed length
AB = √(4^2 + 8^2) = √80
The height from C to line AB must be as long as possible, this is achieved when the slope of the tangent at C is the same as the slope of AB,
dy/dx = 2x
slope of AB = 8/4 = 2
then 2x = 2
x = 1
when x = 1, y = 1^2 = 1
C must be (1,1)
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