Question
My teacher told me to use the distance formula for this problem but I am not sure how to do it...
find the point (s) on the graph of y^2=6x closest to the point (5,0)
find the point (s) on the graph of y^2=6x closest to the point (5,0)
Answers
Reiny
Let the point on the parabola which is closest to (5,0) be P(x,y)
let D be the distance between the two points
D^2 = (x-5)^2 + (y-0)^2
= (x-5)^2 + y^2
= (x-5)^2 + 6x
2D dD/dx = 2(x-5) + 6
for a minimum of D , dD/dx = 0
2(x-5) + 6 = 0
x-5 = -3
x = 2
then y^2 = 6(2)=12
y = ±√12 = ± 2√3
The two closest points are (2, 2√3) and (2, -2√3)
let D be the distance between the two points
D^2 = (x-5)^2 + (y-0)^2
= (x-5)^2 + y^2
= (x-5)^2 + 6x
2D dD/dx = 2(x-5) + 6
for a minimum of D , dD/dx = 0
2(x-5) + 6 = 0
x-5 = -3
x = 2
then y^2 = 6(2)=12
y = ±√12 = ± 2√3
The two closest points are (2, 2√3) and (2, -2√3)
Anonymous
thank u for the help >_< this was so frustrating for me!