,For $y=x^2+1$ and $x=y^2+1$, if PQ is the shortest distance and R,S are points of contact of common tangent, find area PQRS.

Question

From equation, it is evident that curves are symmetric about $y=x$

So I tried writing the equations of tangents $$k=mh +\frac{1}{4m}$$ and $$h=mk-\frac{m^2}{4}$$

Where $(h,k)$ is a point on curve 1

But on plugging $m=1$ the lines turn out to be coincident, and not two separate parallel lines. I was hoping to find $(h,k)$ by solving these two equations but that clearly isn’t working. What should I do?

The two curves are mirror images of each other, reflected in the line $y=x$. So the common tangent and the shortest connecting line segment both have gradient $-1$. — John Bentin, Feb 07 '21 at 13:17
@JohnBentin That’s exactly what I did. I am not able to go beyond that — Aditya, Feb 07 '21 at 13:38
@TonyK: yes, but the one giving a convex triangle is the one in Quanto's answer. The signed area of the other two choices for PQRS is 9/4. — Jan-Magnus Økland, Feb 07 '21 at 14:59

Quanto · Accepted Answer · 2021-02-09T19:08:01.097

2

Per the symmetry with respect to $y=x$, the tangent is $1$ at $P $ and $-1$ at $R$. Then, set $y’=(x^2+1)’=2x =\pm 1$ to obtain $P(\frac12,\frac54)$ and $R(-\frac12,\frac54)$. As a result, $PR=1$, their distances to the symmetry line $y=x$ are respectively $\frac3{4\sqrt2}$ and $\frac7{4\sqrt2}$, and the height of the trapezoid $PQRS $ is $\frac{PR}{\sqrt2} $. Thus, the area is $$ 2\cdot \frac12(\frac3{4\sqrt2}+ \frac7{4\sqrt2}) \frac1{\sqrt2}=\frac54 $$

edited Feb 09 '21 at 19:08

answered Feb 07 '21 at 14:26

Quanto

97,352

I want to know how you managed to find area without writing points R and S – Aditya Feb 08 '21 at 09:37
@Aditya - due to symmetry wrt $y=x$, it is an isosceles trapezoid. So, I only calculated half $\times$ 2 – Quanto Feb 08 '21 at 13:26

score 0 · Answer 2 · answered Feb 07 '21 at 15:53

(Without calculus) The line $y=c-x$ meets $y=x^2+1$ where $x^2+x +1-c+0$. Tangency occurs when this quadratic equation has equal roots, namely when $c=\frac34$. (You may check that the same applies for $x=y^2+1$.) Thus the tangent points are $$R\;(-\tfrac12,\tfrac54)\quad\text{and}\quad S\;(\tfrac54, -\tfrac12).$$

To find the shortest connection between the two curves, note that this is twice the length of the shortest connector from either curve to the line of symmetry $y=x$, say from the point $(t,t)$ on the line to the nearest point on $y=x^2+1$ that lies on the line $y=c-x$ which passes through $(t,t)$, namely the line $y= 2t-x$. This point is given by first solving $2t-x=x^2+1$ for $x$ in terms of $t$: $$x=-\tfrac12+\sqrt{2t-\tfrac34},\qquad y=2t+\tfrac12-\sqrt{2t-\tfrac34}.$$ (The sign of the square root is chosen to get the minimum distance, in the first quadrant.)

We need now to minimize the squared distance of this point from $(t,t)$, which is $$\left(t+\tfrac12-\sqrt{2t-\tfrac34}\right)^2+\left(-t-\tfrac12+\sqrt{2t-\tfrac34}\right)^2=2\left(t+\tfrac12-\sqrt{2t-\tfrac34}\right)^2.$$ To minimize this, write $t=\tfrac12u^2+\tfrac38,$ so that the quantity to be minimized (after dropping unnecessary features) becomes $$\tfrac12u^2+\tfrac78-u=\tfrac12(u-1)^2 +\tfrac38.$$ Thus $u=1$ and $t=\frac78$. Hence the required points on $y=x^2+1$ and $x=y^2+1$ respectively are $$P\;(\tfrac12,\tfrac54)\quad\text{and}\quad Q\;(\tfrac54,\tfrac12).$$ The midpoints of $PQ$ and $RS$ are $(\frac78,\frac78)$ and $(\frac38,\frac38)$, and so the distance between them is $\frac12\surd2$. Also, $|PQ|=\frac34\surd2$ and $|RS|=\frac74\surd2$. Therefore the area of $PQRS$ is $\frac12\surd2\cdot\frac12(\frac34\surd2+\frac74\surd2)$ or $$\frac54.$$

,For $y=x^2+1$ and $x=y^2+1$, if PQ is the shortest distance and R,S are points of contact of common tangent, find area PQRS.

2 Answers2