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the question ask, to find the point on the graph of the curve given by y^2 = 2x that is closest to point (1,4) stuck on this one... does this have to do with parametric equations and taking the derivative of y^2 = 2x to find slope... a little stuck
My answer was modified slightly, but it was correct.
If you had any trouble comprehending my 1st answer, here it is revised.
I am positive my 2nd answer looks right.
If we are given y = y0, then x0 = (y0)²/2.
The distance is given by s² = (x-1)² + (y-4)².
The derivative with x as a function of y is 0 = 2(x-1)(dx/dy) + 2(y-4).
Dividide both sides by 2, giving 0 = (x-1)(dx/dy) + y-4.
Add 4-y to both sides, getting 4-y = (x-1)(dx/dy).
Since (x0,y0) is on the line with that slope, we can put in what y0 for y and x0 for x.
Note that x0 was given at the end of the first paragraph. This gives 4 - y0 = ((y0)²/2 - 1)(dx/dy).
Multiply both sides by 2 and get 8 - 2y0 = ((y0)² - 2)(dx/dy).
Divide both sides by ((y0)² - 2). This gives (8-2•y0)/((y0)²-2) = dx/dy.
To minimize it, we set the derivative = to 0 and not that the only thing that matters is the numerator, for the denominator can never make a fraction 0. Thus, we have 8 - 2•y0 = 0, or 8 = 2•y0, or 4 = y0.
Since y0² = 2•x0, then 4² = 16 = 2•x0, so 8 = x0.
Thus, the point on the curve that is closest is (8,4).
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