151. The contact of plane curves

1. Let $\varphi (x)=ax+b$, so that $y=\varphi (x)$ is a straight line. The conditions for contact at the point for which $x=\xi$ are $f(\xi )=a\xi +b$, $f^{\prime }(\xi )=a$. If we determine $a$ and $b$ so as to satisfy these equations we find $a=f^{\prime }(\xi )$, $b=f(\xi )–\xi f^{\prime }(\xi )$, and the equation of the tangent to $y=f(x)$ at the point $x=\xi$ is $$y=x{f}^{\prime }(\xi )+\{f(\xi )–\xi f^{\prime }(\xi )\},$$ or $y–f(\xi )=(x–\xi )f^{\prime }(\xi )$. Cf. Ex. XXXIX. 5.

2. The fact that the line is to have simple contact with the curve completely determines the line. In order that the tangent should have contact of the second order with the curve we must have $f”(\xi )=\varphi ”(\xi )$,  $f”(\xi )=0$. A point at which the tangent to a curve has contact of the second order is called a point of inflexion.

3. Find the points of inflexion on the graphs of the functions $3x^4–6x^3+1$, $2x/(1+x^2)$, $\sin x$, $a{\cos }^{2}x+b{\sin }^{2}x$, $\tan x$, $\arctan x$.

4. Show that the conic $a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$ cannot have a point of inflexion. [Here $ax+hy+g+(hx+by+f)y_1=0$ and $$a+2h{y}_1+b{y}_1^2+(hx+by+f)y_2=0,$$ suffixes denoting differentiations. Thus at a point of inflexion $$a+2h{y}_1+b{y}_1^2=0,$$ or $$a(hx+by+f{)}^2–2h(ax+hy+g)(hx+by+f)+b(ax+hy+g{)}^2=0,$$ or $$(ab–h^2)\{a{x}^2+2hxy+b{y}^2+2gx+2fy\}+a{f}^2–2fgh+b{g}^2=0.$$ But this is inconsistent with the equation of the conic unless $$a{f}^2–2fgh+b{g}^2=c(ab–h^2)$$ or $abc+2fgh–a{f}^2–b{g}^2–c{h}^2=0$; and this is the condition that the conic should degenerate into two straight lines.]

5. The curve $y=(a{x}^2+2bx+c)/(\alpha x^2+2\beta x+\gamma )$ has one or three points of inflexion according as the roots of $\alpha x^2+2\beta x+\gamma =0$ are real or complex.

[The equation of the curve can, by a change of origin (cf. Ex. XLVI. 15), be reduced to the form $$\eta =\xi /(A{\xi }^2+2B\xi +C)=\xi /\{A(\xi –p)(\xi –q)\},$$ where $p$, $q$ are real or conjugate. The condition for a point of inflexion will be found to be ${\xi }^3–3pq\xi +pq(p+q)=0$, which has one or three real roots according as $\{pq(p–q){\}}^2$ is positive or negative,  according as $p$ and $q$ are real or conjugate.]

6. Discuss in particular the curves $y=(1–x)/(1+x^2)$, $y=(1–x^2)/(1+x^2)$, $y=(1+x^2)/(1–x^2)$.

7. Show that when the curve of Ex. 5 has three points of inflexion, they lie on a straight line. [The equation ${\xi }^3–3pq\xi +pq(p+q)=0$ can be put in the form $(\xi –p)(\xi –q)(\xi +p+q)+(p–q{)}^2\xi =0$, so that the points of inflexion lie on the line $\xi +A(p–q{)}^2\eta +p+q=0$ or $A\xi –4(AC–B^2)\eta =2B$.]

8. Show that the curves $y=x\sin x$, $y=(\sin x)/x$ have each infinitely many points of inflexion.

9. Contact of a circle with a curve. Curvature.¹ The general equation of a circle, viz. $$\begin{array}{}\text{(1)}& (x–a{)}^2+(y–b{)}^2=r^2,\end{array}$$ contains three arbitrary constants. Let us attempt to determine them so that the circle has contact of as high an order as possible with the curve $y=f(x)$ at the point $(\xi ,\eta )$, where $\eta =f(\xi )$. We write ${\eta }_1$, ${\eta }_2$ for $f^{\prime }(\xi )$, $f”(\xi )$. Differentiating the equation of the circle twice we obtain $$\begin{array}{}\text{(2)}& (x–a)+(y–b)y_1=0,\end{array}$$$$\begin{array}{}\text{(3)}& 1+y_1^2+(y–b)y_2=0.\end{array}$$

If the circle touches the curve then the equations  and  are satisfied when $x=\xi$, $y=\eta$, $y_1={\eta }_1$. This gives $(\xi –a)/{\eta }_1=-(\eta –b)=r/\sqrt{1+{\eta }_1^2}$. If the contact is of the second order then the equation  must also be satisfied when $y_2={\eta }_2$. Thus $b=\eta +\{(1+{\eta }_1^2)/{\eta }_2\}$; and hence we find $$a=\xi –\frac{{\eta }_1(1+{\eta }_1^2)}{{\eta }_2},b=\eta +\frac{1+{\eta }_1^2}{{\eta }_2},r=\frac{(1+{\eta }_1^2{)}^{3/2}}{{\eta }_2}.$$

The circle which has contact of the second order with the curve at the point $(\xi ,\eta )$ is called the circle of curvature, and its radius the radius of curvature. The measure of curvature (or simply the curvature) is the reciprocal of the radius: thus the measure of curvature is $f”(\xi )/\{1+[f^{\prime }(\xi ){]}^2{\}}^{3/2}$, or $$\frac{d^2\eta }{d{\xi }^2}/\{1+(\frac{d\eta }{d\xi }{)}^2{\}}^{3/2}.$$

10. Verify that the curvature of a circle is constant and equal to the reciprocal of the radius; and show that the circle is the only curve whose curvature is constant.

11. Find the centre and radius of curvature at any point of the conics $y^2=4ax$, $(x/a{)}^2+(y/b{)}^2=1$.

12. In an ellipse the radius of curvature at $P$ is $C{D}^3/ab$, where $CD$ is the semi-diameter conjugate to $CP$.

13. Show that in general a conic can be drawn to have contact of the fourth order with the curve $y=f(x)$ at a given point $P$.

[Take the general equation of a conic, viz. $$a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0,$$ and differentiate four times with respect to $x$. Using suffixes to denote differentiation we obtain $$\begin{array}{rl}ax+hy+g+(hx+by+f)y_1& =0,\\ a+2h{y}_1+b{y}_1^2+(hx+by+f)y_2& =0,\\ 3(h+b{y}_1)y_2+(hx+by+f)y_3& =0,\\ 4(h+b{y}_1)y_3+3b{y}_2^2+(hx+by+f)y_4& =0.\end{array}$$ If the conic has contact of the fourth order, then these five equations must be satisfied by writing $\xi$, $\eta$, ${\eta }_1$, ${\eta }_2$, ${\eta }_3$, ${\eta }_4$, for $x$, $y$, $y_1$, $y_2$, $y_3$, $y_4$. We have thus just enough equations to determine the ratios $a:b:c:f:g:h$.]

14. An infinity of conics can be drawn having contact of the third order with the curve at $P$. Show that their centres all lie on a straight line.

[Take the tangent and normal as axes. Then the equation of the conic is of the form $2y=a{x}^2+2hxy+b{y}^2$, and when $x$ is small one value of $y$ may be expressed (Ch. V, Misc. Ex. 22) in the form $$y=\frac{1}{2}a{x}^2+(\frac{1}{2}ah+{ϵ}_x)x^3,$$ where ${ϵ}_x\to 0$ with $x$. But this expression must be the same as $$y=\frac{1}{2}f”(0)x^2+\{\frac{1}{6}f{”}^{\prime }(0)+{ϵ}_x^{\prime }\}{x}^3,$$ where ${ϵ}_x^{\prime }\to 0$ with $x$, and so $a=f”(0)$, $h=f{”}^{\prime }(0)/3f”(0)$, in virtue of the result of Ex. LV. 15. But the centre lies on the line $ax+hy=0$.]

15. Determine a parabola which has contact of the third order with the ellipse $(x/a{)}^2+(y/b{)}^2=1$ at the extremity of the major axis.

16. The locus of the centres of conics which have contact of the third order with the ellipse $(x/a{)}^2+(y/b{)}^2=1$ at the point $(a\cos \alpha ,b\sin \alpha )$ is the diameter $x/(a\cos \alpha )=y/(b\sin \alpha )$. [For the ellipse itself is one such conic.]