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

If this part uses an earlier result, revisit (i) Optimisation: vertical gap or open all working together.

(ii)

Optimisation: horizontal gap

PP on CC and QQ on LL have the same yy-coordinate. Find PP and QQ that give the shortest distance PQPQ.

Explore this idea

Hint

  1. Write xx in terms of yy for each: PP has x=yx=\sqrt y (the nearer branch) and QQ has x=2(y+2)x=2(y+2).

  2. Minimise PQ=2(y+2)−yPQ=2(y+2)-\sqrt y.

Answer · (ii)

P=(1/4,1/16)P=(1/4,1/16); Q=(33/8,1/16)Q=(33/8,1/16).

Minimum PQ=31/8.

Explore the idea

Open interactive exploration

The horizontal gap PQ against y

As y increases from 0 the gap first shrinks, then grows. Its gradient is zero at y = 1/16.

Position along the curvey = 0.5
Try an example
Test a critical position

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

f(y)
4.29289
Gradient f′(y)
1.29289
positive

For y>1/16 the horizontal gap grows.

Selected values are within the original drawing windows.

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

The horizontal and vertical gaps are different functions, and here both are smallest at the same point (14,116)\left(\frac14,\frac1{16}\right) on CC.

Ideas to try
Is the minimum at the same yy for both gaps?