Question

Find the point on the parabola 4y= x^2 nearest to the point (1,2).

Answer 1

we need to find a normal to the line that perps to 1,2 then

Answer 2

first; lets define the slope of the line for any given point on it:

4y = x^2

4 y' = 2x

y' = (1/2)x this is tell us the slope of the tangent line at any given x

Answer 3

Can i answer through skype?? If not answer is (3.538584704/2,3.538584704^2/8)

Answer 4

to perp it; we flip and negate it

Answer 5

slope = (1/2)x

perp slope = -2/x

and we want it to run thru the point (1,2); and hopefully im thinking thru this correctly :)

Answer 6

the equation for the the perp line becomes:

y = -2x/x + 2(1)/x + 2

y = -2 + 2/x + 2

y = 2/x .... i got me doubts

Answer 7

so where does:

4y= x^2 and y = 2/x meet by chance?

x^2/4 = 2/x

x^3 = 8

x=2 hmmmm

if im right, and i doubt it; id say at (2,1)

Answer 8

I get (2,1) also

Answer 9

you can also use the distance formula to solve this problem

Answer 10

well, if it isnt right, then at least its a consensus lol

Answer 11

it is correct

Answer 12

\[d^2=(x-1)^2+(y-2)^2=(x-1)^2+(x^2/4-2)^2\]\[(d^2)'=2(x-1)+2(x^2/4-2)\frac{x}{2}=0\]\[2x-2+\frac{x^3}{4}-2x=0\quad\Rightarrow x^3=8\quad,\quad x=2\quad,\quad y=1\]