Parabola to a point

Calculus Level 2

Find the shortest distance from the parabola x 2 = 20 y x^2 = 20y to the point ( 15 , 20 ) (15,20) . Give your answer correct to two decimal places.


The answer is 4.45.

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2 solutions

Chew-Seong Cheong
Jan 31, 2020

Let the fixed point be Q ( 15 , 20 ) Q(15, 20) and point P ( x , y ) P(x,y) on the parabola where P Q PQ is the shortest. Then P Q PQ must be perpendicular to the tangent at point P P . Since the gradient at point P P is given by d y d x = x 10 \dfrac {dy}{dx} = \dfrac x{10} . Then the gradient of line P Q PQ is 10 x -\dfrac {10}x and the following equation is true:

y 20 x 15 = 10 x Note that y = x 2 20 x 2 20 20 = 150 x 10 x 3 200 x 3000 = 0 \begin{aligned} \frac {\blue y-20}{x-15} & = - \frac {10}x & \small \blue{\text{Note that }y = \frac {x^2}{20}} \\ \frac {x^2}{20} - 20 & = \frac {150}x - 10 \\ \implies x^3 - 200x - 3000 & = 0 \end{aligned}

We can use Cardano's method to solve the cubic equation as follows:

x 3 200 x 3000 = 0 Let x = u + v u 3 + v 3 + ( 3 u v 200 ) ( u + v ) 3000 = 0 Assume 3 u v 200 = 0 u 3 + v 3 = 3000 u 3 v 3 = 8000000 27 \begin{aligned} x^3 - 200x - 3000 & = 0 & \small \blue{\text{Let }x = u+v} \\ u^3 + v^3 + (3uv-200)(u+v) - 3000 & = 0 & \small \blue{\text{Assume }3uv-200=0} \\ u^3 + v^3 & = 3000 & \small \blue{\implies u^3v^3 = \frac {8000000}{27}} \end{aligned}

By Vieta's formula , u 3 u^3 and v 3 v^3 are roots of the following quadratic equation.

w 2 3000 w + 8000000 27 = 0 \begin{aligned} w^2 - 3000 w + \frac {8000000}{27} & = 0 \end{aligned}

w { 2897.749514 u 14.25674168 102.2504861 v 4.676150286 \implies w \approx \begin{cases} 2897.749514 & \implies u \approx 14.25674168 \\ 102.2504861 & \implies v \approx 4.676150286 \end{cases}

x = u + v 18.93289196 y 17.9227199 \implies x = u+v \approx 18.93289196 \implies y \approx 17.9227199 and P Q ( 15 18.93289196 ) 2 + ( 20 17.9227199 ) 2 4.45 PQ \approx \sqrt{(15-18.93289196)^2+(20-17.9227199)^2} \approx \boxed{4.45} .

Let us take a point ( h , h 2 20 ) (h, \dfrac{h^2}{20}) on the parabola. Then the square of the distance of this point from ( 15 , 20 ) (15,20) is ( h 15 ) 2 + ( h 2 20 20 ) 2 (h-15)^2+(\dfrac{h^2}{20}-20)^2 . This attains a minimum when h 3 200 h 3000 = 0 h^3-200h-3000=0 or h = 18.93289... h=18.93289... . The minimum distance is 4.4477782981... \boxed {4.4477782981...}

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