Question #280718

Minimum distance. Find the minimum distance from a point on the

positive x-axis (a, 0) to the parabola y^2 = 8x.


1
Expert's answer
2021-12-20T05:23:08-0500

Given parabola y² = 8x


and the point (a,0).


Let P(x, y) be any point on the parabola.


d=(xa)2+y2d = \sqrt{(x − a ) ² + y²}


D=d2=(xa)2+y2D=d² = (x-a)² + y²


D(x)=(xa)2+8xD(x)= (x − a)² +8x

Differentiate with respect to x


D(x)=2(xa)+8D'(x) = 2(x-a)+8


D(x)=0D'(x) = 0


2(xa)+8=02(x-a)+8=0


xa=4x-a=-4


x=4+ax = -4+a

D(x)=2(xa)+8D'(x) = 2(x-a)+8


Differentiate with respect to x


D(x)=2xD''(x) = 2x


D(4+a)=2(4+a)=8+aD''(-4+a)= 2(-4+a)= -8+a


If a > 8 then D"(-4+a) >0


The minimum distance is

(4+aa)2+8(4+a)\sqrt{(-4+a-a)² +8(-4+a)}


=1612+8a\sqrt{16-12+8a}


= 4+8a\sqrt{4+8a}



Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS