Shortest Distance on a Parabola: Point to Point Calculation

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2. Find the shortest distance from the point (1,4) to a point on the parabola y^2 = 2x

Not really sure what to do here next. I'd imagine i might have to fuse implicit differentiation? But not really sure.
 
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1) solve for x
2) find tangent to curve at x
3) use point slope formula
4) solve for x

your ans should be sqrt(5)
 
I'm kind of confused on step 2. How do you find the tangent?
 
you know the x point so find its derivative and use the point slope formula and solve for x
 
I still don't really understand. Do you think you can show it?
 
There are multiple ways to solve this problem. One way is to think about this as an minimization problem. Hint: Let (x,y) be the point on the parabola closest to (1,4) and write out the formula for the distance between these two points. Next, use the relationship between x and y to write the formula in terms of one variable. Now what do you think you should do?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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