Question

Find the points on the parabola y=x2 that are closest to the point (0,5).

Answer 1

ok so the length of the line from (0,5) to a point on the parabola x² becomes a function (rise/run): (5-x²)/(0-x)=5-x²/x

Answer 2

that can be differentiated then to find the maximum value point on the f(x)=5-x²/x: f'(x)=-x²-5/x²

Answer 3

Let the required point be (p, p^2).

Distance from P to (0, 5) squared
D^2 = (p - 0))^2 + (p^2 - 5)^2 = (p)^2 + (p^2 - 5)^2

Multiply out and differentiate to get 2D*dD/dp = f(p)
For minimum distance solve f(p) = 0.

Answer 4

http://www.mesacc.edu/~marfv02121/readings/nearest_point/index.html

Answer 5

the points i got were:
\[x = (\sqrt{2} \times-3) / 2 or x = (2√×3)/2\]

Answer 6

9/2, (-3 x sqrt2)/2

Answer 7

@Jack1 can you clarify 2D*dD/dp ? I didn't follow

Answer 8

\[d^2=x^2+(x^2-5)^2\]
\[d^2=x^2+x^4-10x^2+25=x^4-9x^2+25\]

Answer 9

derivative is \(4x^3-18x\) which is zero is \(x=0\) or if \(4x^2-18=0\)

Answer 10

second one gives \(x^2=\frac{9}{2}\)

Answer 11

@satellite73 why is it ok to use the square of the distance?

Answer 12

because the ratios of x y and d remain the same ...

Answer 13

since d = sqrt(x^2+y^2, then d^2 = x^2 + y^2

Answer 14

my idea is to take the derivative; 2x, as the general slope of a tangent line; therefore we want the perp line (for closest point) to be -1/2x for any point (a,b) on the function
\[y=-\frac{1}{2a}(x-a)+b\]to make sure the line has the stated point (0,5), plug it in
\[0=-\frac{1}{2a}(5-a)+b\]
\[0=-\frac{5}{2a}+\frac52+b\]
\[b=\frac{5}{2a}-\frac52\]set y=x^2 to a,b and equate
\[a^2-b=\frac{5}{2a}-\frac52-b\]

Answer 15

im sure i messed that up along the way, but it was a good effort lol

Answer 16

yes, i used (0,5) as (y,x) instead of (x,y) :/

Answer 17

\[5=-\frac{1}{2a}(0-a)+b\]
\[5=\frac12+b\]
\[b=4.5\]therefore y = 4.5 when x = sqrt(4.5) is my guess

Answer 18

@amistre64 what does the solution x=0 mean when the point (0,y) is clearly not one of the nearest points?

Answer 19

ah it's a local maximum?

Answer 20

x=0 is a line

Answer 21

x=0 is also known as the y axis, they represent the same graphical line

Answer 22

I mean, if the derivative has three minimum or maximum points (-sqrt(9/2), 0, sqrt(9/2)), how do I check which ones they are? I know intuitively

Answer 23

the derivative =0, or undefined, give us critical values that we need to check by plugging those critical values back into the original setup.

we can also use another derivative and check for sign changes if need be.

Answer 24

btw, the derivative of x^2 is just 2x, and has only one critical point, 2x = 0 when x=0

Answer 25

the derivative of 2x is 2, which is always positive and tells us the slope is increasing as x approaches infinity