Find the shortest distance from the parabola x 2 = 2 0 y to the point ( 1 5 , 2 0 ) . Give your answer correct to two decimal places.
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Let us take a point ( h , 2 0 h 2 ) on the parabola. Then the square of the distance of this point from ( 1 5 , 2 0 ) is ( h − 1 5 ) 2 + ( 2 0 h 2 − 2 0 ) 2 . This attains a minimum when h 3 − 2 0 0 h − 3 0 0 0 = 0 or h = 1 8 . 9 3 2 8 9 . . . . The minimum distance is 4 . 4 4 7 7 7 8 2 9 8 1 . . .
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Let the fixed point be Q ( 1 5 , 2 0 ) and point P ( x , y ) on the parabola where P Q is the shortest. Then P Q must be perpendicular to the tangent at point P . Since the gradient at point P is given by d x d y = 1 0 x . Then the gradient of line P Q is − x 1 0 and the following equation is true:
x − 1 5 y − 2 0 2 0 x 2 − 2 0 ⟹ x 3 − 2 0 0 x − 3 0 0 0 = − x 1 0 = x 1 5 0 − 1 0 = 0 Note that y = 2 0 x 2
We can use Cardano's method to solve the cubic equation as follows:
x 3 − 2 0 0 x − 3 0 0 0 u 3 + v 3 + ( 3 u v − 2 0 0 ) ( u + v ) − 3 0 0 0 u 3 + v 3 = 0 = 0 = 3 0 0 0 Let x = u + v Assume 3 u v − 2 0 0 = 0 ⟹ u 3 v 3 = 2 7 8 0 0 0 0 0 0
By Vieta's formula , u 3 and v 3 are roots of the following quadratic equation.
w 2 − 3 0 0 0 w + 2 7 8 0 0 0 0 0 0 = 0
⟹ w ≈ { 2 8 9 7 . 7 4 9 5 1 4 1 0 2 . 2 5 0 4 8 6 1 ⟹ u ≈ 1 4 . 2 5 6 7 4 1 6 8 ⟹ v ≈ 4 . 6 7 6 1 5 0 2 8 6
⟹ x = u + v ≈ 1 8 . 9 3 2 8 9 1 9 6 ⟹ y ≈ 1 7 . 9 2 2 7 1 9 9 and P Q ≈ ( 1 5 − 1 8 . 9 3 2 8 9 1 9 6 ) 2 + ( 2 0 − 1 7 . 9 2 2 7 1 9 9 ) 2 ≈ 4 . 4 5 .