Answer to Question #259062 in Calculus for Chay

Question

Find the minimum distance from the point (4, 2) to the parabola y²=8x.

Answer 1

Solution:

Given, $y^2=8x$ and point is (4, 2).

Let y=a and put in above equation.

$a^2=8x \\\Rightarrow x=\dfrac{a^2}8$

so, the point $(\dfrac{a^2}8,a)$ lies on the curve.

Let the distance between $(\dfrac{a^2}8,a)$ and (4, 2) be given by D.

Then,

$D^2=(\dfrac{a^2}8-4)^2+(a-2)^2 \\ \Rightarrow D^2=\dfrac1{64}(a^2-32)^2+(a-2)^2\ ...(i)$

Differentiating both sides of (i) w.r.t $a$,

$2D.D'=\dfrac1{32}(a^2-32)(2a)+2(a-2)$

We need minimum distance, so put $D'=0$

$0=\dfrac1{32}(a^2-32)(2a)+2(a-2) \\ \Rightarrow 0=\dfrac1{16}a(a^2-32)+2(a-2) \\ \Rightarrow 0=a(a^2-32)+32(a-2) \\ \Rightarrow 0=a^3-32a+32a-64 \\ \Rightarrow 0=a^3-64 \\ \Rightarrow a=4$

Put this value in (i).

$D^2=\dfrac1{64}(4^2-32)^2+(4-2)^2 \\ \Rightarrow D^2=\dfrac1{64}(-16)^2+(2)^2 \\ \Rightarrow D^2=\dfrac1{64}(256)+4 \\ \Rightarrow D^2=4+4=8 \\ \Rightarrow D=2\sqrt2\ units$