Mathalicious

[Inside Out]

1. Find the point of the parabola y=x² that is closest to the point (3,0)

[Water]

Think Tangent, Pythagoras Theorem, and you'll get the answer.

[Inside Out]

so the derivative would be 2x... what do i do with the rest

[xZero]

i prefer using , sub x1 and y1 with (3,0) and x2 y2 with (x,x^2)

[Water]

I don't think you need Distance rule in this.



After you have your derivative 2x, imagine a triangle, right angle, and perpendicular, to get your shortest side.


The perpendicular of 2x would be -1/2x

Equation for the perpendicular (3,0)

y = -x/2 + 3/2


You equate them together, to find point of intersection that creates the triangle to (3,0)


2x = -x/2 + 3/2

x = 1

Sub into x^2

(1,1)



PS: made a mistake in my working out before: correction

[xZero]

You can, diff the distance rule and solve for x, which will get you x=1

[Inside Out]

how do u know that the shortest side will be at right angles with the tangent?

[brightsky]

logic. draw a point and a line. now draw the line perpendicular to the first, passing through the point. now draw random lines joining the point and the first line which are not perpendicular to it. by pythagoras, said lines would effectively be the hypotenuse, which is longer than the original perpendicular line.