Vertical and horizontal gaps to a parabola · source reference

Printed question

The curve $C$ and the line $L$ have equations $y=x^2$ and $y=\frac12x-2$ respectively.

(i) The point $A$ on $C$ and the point $B$ on $L$ are such that they have the same $x$-coordinate. Find the coordinates of $A$ and $B$ that gives the shortest distance $AB$. [3]

(ii) The point $P$ on $C$ and the point $Q$ on $L$ are such that they have the same $y$-coordinate. Find the coordinates of $P$ and $Q$ that gives the shortest distance $PQ$. [3]

(iii) Find the exact area of the polygon formed by joining the points found in (i) and (ii). [2]

(iv) A variable point on $C$ with coordinates $(s,s^2)$ starts from origin $O$ and moves along the curve with $s$ increasing at a rate of 2 units/s. Find the rate of change of the area bounded by the curve, the $y$-axis and the line $y=s^2$, at the instant when $s=\sqrt2$. [4]

[2017/ACJC/Prelim/I/8]

Printed working

(i) Let $V$ be $AB$. $V=y_1-y_2=x^2-(x/2-2)=x^2-x/2+2$, $dV/dx=2x-1/2$. For maximum/minimum, $dV/dx=0\Rightarrow x=1/4$. $d^2V/dx^2=2>0$, so minimum there. On $C$, $y=(1/4)^2=1/16$; on $L$, $y=(1/2)(1/4)-2=-15/8$.

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

Alternative printed solution: $A=(x,x^2)$, $B=(x,x/2-2)$.

$$AB=x^2-(x/2-2)=(x-1/4)^2-1/16+2=(x-1/4)^2+31/16.$$

Thus minimum $AB=31/16$, attained at $x=1/4$.

(ii) Let $H$ be $PQ$. $H=x_2-x_1=2(y+2)-\sqrt y$. $dH/dy=2-\frac12y^{-1/2}$. Set zero: $2-\frac12y^{-1/2}=0\Rightarrow2=\frac12y^{-1/2}\Rightarrow y=4^{-2}=1/16$.

$$\frac{d^2H}{dy^2}=\frac14y^{-3/2};\quad y=1/16\Rightarrow H''=\frac14(1/16)^{-3/2}=16>0.$$

Minimum at $y=1/16$. On $C$, $x=\sqrt{1/16}=1/4$; on $L$, $x=2(1/16+2)=33/8$. The source main working prints $x=2(1/16)+2=33/8$; this intermediate $+2$ is a typographical error, and must be $+4$. Thus $P=(1/4,1/16)$, $Q=(33/8,1/16)$.

Alternative printed solution: $P=(\sqrt y,y)$, $Q=(2y+4,y)$.

$$PQ=2y+4-\sqrt y=2(\sqrt y)^2-\sqrt y+4=2(\sqrt y-1/4)^2-2(1/16)+4=2(\sqrt y-1/4)^2+62/16.$$

The source has a typesetting slip in the subtraction, displaying $-2(1/16)^2$ on one line; the correct completed-square constant is $-2(1/4)^2=-1/8$, as in the final $62/16$. Minimum $PQ=62/16$ at $\sqrt y=1/4$, i.e. $y=1/16$.

(iii) $A$ and $P$ coincide, so polygon is a triangle. $AB=1/16-(-15/8)=31/16$; $PQ=33/8-1/4=31/8$.

Area $=\frac12\times\frac{31}{16}\times\frac{31}8=\frac{961}{256}$ square units.

(iv)

$$A=\int_0^{s^2}x\,dy=\int_0^{s^2}\sqrt y\,dy=\left[\frac{y^{3/2}}{3/2}\right]_0^{s^2}=\frac23s^3.$$ $$\frac{dA}{dt}=\frac{dA}{ds}\frac{ds}{dt}=2s^2(2)=4s^2.$$

At $s=\sqrt2$, $dA/dt=4(\sqrt2)^2=8$ units$^2$/s.

Alternative printed solution: area is rectangle minus area below curve:

$$A=s\times s^2-\int_0^s y\,dx=s^3-\int_0^s x^2\,dx=s^3-\left[\frac{x^3}3\right]_0^s=\frac23s^3.$$

Again $dA/dt=(dA/ds)(ds/dt)=2s^2(2)=4s^2$, giving $8$ units$^2$/s at $s=\sqrt2$.

Teacher notes

Use $x_1=\sqrt y$, not $-\sqrt y$, because the shortest distance is wanted: use the nearer branch.

Note that $A$ and $P$ are the same point, so the polygon has three distinct vertices and its area is a triangle area.

Answer key

(i) $A(1/4,1/16)$, $B(1/4,-15/8)$. (ii) $P(1/4,1/16)$, $Q(33/8,1/16)$. (iii) $961/256$ units$^2$. (iv) $8$ units$^2$/s.

Native diagrams