Vertical and horizontal gaps to a parabola

The curve CC and the line LL have equations y=x2y=x^2 and y=12x−2y=\frac12x-2 respectively.

(i) The point AA on CC and the point BB on LL are such that they have the same xx-coordinate. Find the coordinates of AA and BB that gives the shortest distance ABAB. marks

(ii) The point PP on CC and the point QQ on LL are such that they have the same yy-coordinate. Find the coordinates of PP and QQ that gives the shortest distance PQPQ. marks

(iii) Find the exact area of the polygon formed by joining the points found in (i) and (ii). marks

(iv) A variable point on CC with coordinates (s,s2)(s,s^2) starts from origin OO and moves along the curve with ss increasing at a rate of 2 units/s. Find the rate of change of the area bounded by the curve, the yy-axis and the line y=s2y=s^2, at the instant when s=2s=\sqrt2. marks

[2017/ACJC/Prelim/I/8]

Total marks: 12

Solution

(i)

Optimisation: vertical gap

C:y=x2C:y=x^2 and L:y=12x−2L:y=\frac12x-2. AA on CC and BB on LL have the same xx-coordinate. Find AA and BB that give the shortest distance ABAB.

Explore this idea

Hint

  1. For the same xx, AB=x2−(x2−2)AB=x^2-\left(\frac x2-2\right).

  2. Minimise this function of xx.

Answer · (i)

A=(1/4,1/16)A=(1/4,1/16); B=(1/4,−15/8)B=(1/4,-15/8).

Minimum AB=31/16.

Explore the idea

Open interactive exploration

The vertical gap AB against x

The gap is smallest where its gradient is zero: x = 1/4.

Position along the curvex = -1
Try an example
Test a critical position

Buttons use full-precision reviewed positions; displayed coordinates are rounded.

f(x)
3.5
Gradient f′(x)
-2.5
negative

For x<1/4 the vertical gap shrinks as x increases.

Selected values are within the original drawing windows.

  • Vertical gap AB
  • Tangent
  • Gradient, and where the gap grows
  • Join P to Q; the dashed legs show Δx and Δy
Key idea

The gap is a quadratic in xx; its vertex is at x=14x=\frac14.

Ideas to try
Find the xx with the smallest gap.