MHB Shortest distance between (18,0) and y=x^2

[karush]

What is the shortest distance between $(18,0)$ and $y=x^2$
I presume this could be solved by slopes but couldn't get the formula set up

 

[kaliprasad]

What is the shortest distance between $(18,0)$ and $y=x^2$
I presume this could be solved by slopes but couldn't get the formula set up

take any point say (x,x^2) on the graph
find the distance from (18 .0) that is $\sqrt{(18-x)^2 + (x^2)^2}$
minimize it or minimize $(18-x)^2 + x^4$

now u can proceed with help of calculus

 

[karush]

Since $y=x^2$ then $y'=2x=m$

So
$${x}^{2 }=\frac{1}{-2x}\left(x-18\right)$$

From which we get $(2,4)$
$$\sqrt{\left(18-2\right)^2+\left(4-0\right)^2}=4\sqrt{17}\approx 16.5=d$$