Answer to Question #280718 in Calculus for ash

Question

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

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

Answer 1

Given parabola y² = 8x

and the point (a,0).

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

$d = \sqrt{(x − a ) ² + y²}$

$D=d² = (x-a)² + y²$

$D(x)= (x − a)² +8x$

Differentiate with respect to x

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

$D'(x) = 0$

$2(x-a)+8=0$

$x-a=-4$

$x = -4+a$

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

Differentiate with respect to x

$D''(x) = 2x$

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

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

The minimum distance is

$\sqrt{(-4+a-a)² +8(-4+a)}$

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

= $\sqrt{4+8a}$