Chapter 4 Tangents and Normals-II

4.1 Question 11

• Prove that the sum of the intercepts of the tangent to the curve √x+√y=√a$\sqrt{x}+\sqrt{y}=\sqrt{a}$ upon the coordinate axes is constant.

Differentiating w.r.t x$x$:

dydx=−fxfydydx=−12√x12√y=−√y√x$$\begin{array}{cc}\frac{dy}{dx}& =-\frac{f_x}{f_y}\\ \frac{dy}{dx}& =-\frac{\frac{1}{2\sqrt{x}}}{\frac{1}{2\sqrt{y}}}\\ & =-\frac{\sqrt{y}}{\sqrt{x}}\end{array}$$

The equation of the tangent to the curve is:

Y−y=−√y√x(X−x)X√x+Y√y=√x+√yX√x+Y√y=√aX√x√a+Y√y√a=1$$\begin{array}{cc}Y-y& =-\frac{\sqrt{y}}{\sqrt{x}}(X-x)\\ \frac{X}{\sqrt{x}}+\frac{Y}{\sqrt{y}}& =\sqrt{x}+\sqrt{y}\\ \frac{X}{\sqrt{x}}+\frac{Y}{\sqrt{y}}& =\sqrt{a}\\ \frac{X}{\sqrt{x}\sqrt{a}}+\frac{Y}{\sqrt{y}\sqrt{a}}& =1\end{array}$$

This equation is in double intercept form. So, the x$x$-intercept is √x√a$\sqrt{x}\sqrt{a}$ and y$y$-intercept is √y√a$\sqrt{y}\sqrt{a}$. As per question, we have to prove sum of these intercepts is constant i.e √x√a+√y√a=constant$\sqrt{x}\sqrt{a}+\sqrt{y}\sqrt{a}=\text{constant}$.

√x√a+√y√a=√a(√x+√y)=√a√a=a, which is indeed a constant.$$\begin{array}{cc}\sqrt{x}\sqrt{a}+\sqrt{y}\sqrt{a}& =\sqrt{a}(\sqrt{x}+\sqrt{y})\\ & =\sqrt{a}\sqrt{a}\\ & =a\text{, which is indeed a constant.}\end{array}$$

• Show that the portion of the tangent at any point on the curve x=acos3θ,y=asin3θ$x=a{\cos }^3\theta ,y=a{\sin }^3\theta$ intercepted between the axes is of constant length.

For this,

dxdθ=−3acos2θsinθdydθ=3asin2θcosθdydx=dydθdxdθ=−3asin2θcosθ3acos2θsinθdydx=−tanθ$$\begin{array}{cc}\frac{dx}{d\theta }& =-3a{\cos }^2\theta \sin \theta \\ \frac{dy}{d\theta }& =3a{\sin }^2\theta \cos \theta \\ \frac{dy}{dx}& =\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}\\ & =-\frac{3a{\sin }^2\theta \cos \theta }{3a{\cos }^2\theta \sin \theta }\\ \frac{dy}{dx}& =-\tan \theta \end{array}$$

The equation of tangent is:

Y−y=−tanθ(X−x)tanθX+Y=xtanθ+y=acos3θ×sinθcosθ+asin3θ=asinθ(cos2θ+sin2θ)tanθX+Y=asinθ$$\begin{array}{cc}Y-y& =-\tan \theta (X-x)\\ \tan \theta X+Y& =x\tan \theta +y\\ & =a{\cos }^3\theta \times \frac{\sin \theta }{\cos \theta }+a{\sin }^3\theta \\ & =a\sin \theta ({\cos }^2\theta +{\sin }^2\theta )\\ \tan \theta X+Y& =a\sin \theta \end{array}$$

Dividing both sides by asinθ$a\sin \theta$,

Xacosθ+Yasinθ=1$$\begin{array}{cc}\frac{X}{a\cos \theta }+\frac{Y}{a\sin \theta }& =1\end{array}$$

This equation is in double intercept form. So, the tangent meets the x$x$-axis at X=acosθ$X=a\cos \theta$ and the y$y$-axis at Y=asinθ$Y=a\sin \theta$.

The portion or the length of tangent between two coordinate axes is:

(x-intercept)2+(y-intercept)2a2cos2θ+a2sin2θa2(cos2θ+sin2θ)a2×1a2$$\begin{array}{c}(\text{x-intercept}{)}^2+(\text{y-intercept}{)}^2\\ a^2{\cos }^2\theta +a^2{\sin }^2\theta \\ a^2({\cos }^2\theta +{\sin }^2\theta )\\ a^2\times 1\\ a^2\end{array}$$

which is of constant length.

• In a catenary y=c coshxc$y=c\text{cosh}\frac{x}{c}$, show that the length of the perpendicular from the foot of the ordinate on the tangent is of constant length and that the length of the normal at any point is y2c$\frac{y^2}{c}$.

In physics and geometry, a catenary is the curve that an idealized hanging chain or cable assumes under its own weight when supported only at its ends. The catenary curve has a U-like shape, superficially similar in appearance to a parabolic arch, but it is not a parabola.

The plotting of the graph reveals this curve:

In the above figure, MN is the ordinate of point M, where the tangent meets the curve. Lets assume the tangent meets the curve at point M with coordinate (x,y)$(x,y)$. So point N is the foot of the ordinate MN. We have to find out length of the perpendicular from the foot of the ordinate on the tangent i.e. ON and length of normal MP between the point P and x$x$-axis.

Differentiating w.r.t x$x$:

dydx=c sinhxc×1c= sinhxc$$\begin{array}{cc}\frac{dy}{dx}& =c\text{sinh}\frac{x}{c}\times \frac{1}{c}\\ & =\text{sinh}\frac{x}{c}\end{array}$$

The equation of the tangent is:

Y−y= sinhxc(X−x) sinhxcX−Y+(y−x sinhxc)=0$$\begin{array}{cc}Y-y& =\text{sinh}\frac{x}{c}(X-x)\\ \text{sinh}\frac{x}{c}X-Y+(y-x\text{sinh}\frac{x}{c})& =0\end{array}$$

The coordinate of point N is (x,0)$(x,0)$. What we know is the distance of a perpendicular line from a point (x1,y1)$(x_1,y_1)$ is:

|Ax1+By1+C|√A2+B2$$\begin{array}{c}\frac{|A{x}_1+B{y}_1+C|}{\sqrt{A^2+B^2}}\end{array}$$

So in our case the length of ON is thus:

ON =| sinhxcx+B×0+y−x sinhxc|√( sinhxc)2+(−1)2=y√( coshxc)2=y coshxc=yyc=c, which is a constant.$$\begin{array}{cc}\text{ON}& =\frac{|\text{sinh}\frac{x}{c}x+B\times 0+y-x\text{sinh}\frac{x}{c}|}{\sqrt{(\text{sinh}\frac{x}{c}{)}^2+(-1{)}^2}}\\ & =\frac{y}{\sqrt{(\text{cosh}\frac{x}{c}{)}^2}}\\ & =\frac{y}{\text{cosh}\frac{x}{c}}\\ & =\frac{y}{\frac{y}{c}}=c\text{, which is a constant.}\end{array}$$

Thus the length of the perpendicular from the foot of the ordinate on the tangent is of constant length c$c$.

The length of the normal MP is given by:

MP =y√1+y′2=c coshxc√1+( sinhxc)2=c coshxc coshxc=c×yc×yc=y2c$$\begin{array}{cc}\text{MP}& =y\sqrt{1+y^{\prime 2}}\\ & =c\text{cosh}\frac{x}{c}\sqrt{1+(\text{sinh}\frac{x}{c}{)}^2}\\ & =c\text{cosh}\frac{x}{c}\text{cosh}\frac{x}{c}\\ & =c\times \frac{y}{c}\times \frac{y}{c}\\ & =\frac{y^2}{c}\end{array}$$

Hence proved.

4.2 Question 12

4.2.1 Find the angle between the pair of curves:

• x2−y2=a2$x^2-y^2=a^2$ and x2+y2=√2a2$x^2+y^2=\sqrt{2}{a}^2$

The plotting of the graph reveals this curve:

The angle between two curves is:

tanθ=±m1−m21+m1m2$$\begin{array}{cc}\tan \theta & =\pm \frac{m_1-m_2}{1+m_1{m}_2}\end{array}$$

where m1$m_1$ and m2$m_2$ are the slope of two curves.

The two curves intersect at point x=√a2(√2+1)2$x=\sqrt{\frac{a^2(\sqrt{2}+1)}{2}}$ and y=√a2(√2−1)2$y=\sqrt{\frac{a^2(\sqrt{2}-1)}{2}}$. In our case, for first curve:

dydx=−fxfy=−2x−2y=xy$$\begin{array}{cc}\frac{dy}{dx}& =-\frac{f_x}{f_y}\\ & =-\frac{2x}{-2y}\\ & =\frac{x}{y}\end{array}$$

For second curve,

dydx=−fxfy=−2x2y=−xy$$\begin{array}{cc}\frac{dy}{dx}& =-\frac{f_x}{f_y}\\ & =-\frac{2x}{2y}\\ & =-\frac{x}{y}\end{array}$$

So angle between the two curves is,

tanθ=±m1−m21+m1m2=±xy+xy1−xyxy=±2xyy2−x2$$\begin{array}{cc}\tan \theta & =\pm \frac{m_1-m_2}{1+m_1{m}_2}\\ & =\pm \frac{\frac{x}{y}+\frac{x}{y}}{1-\frac{x}{y}\frac{x}{y}}\\ & =\pm \frac{2xy}{y^2-x^2}\end{array}$$

Substituting for values with point of intersection:

tanθ=±2√a2(√2+1)2√a2(√2−1)2(√a2(√2−1)2)2−(√a2(√2+1)2)2tanθ=±a2a2(√2−1−√2−1)2=±a2−a2tanθ=±−1tanθ=±1θ= π/4 and 3π/4$$\begin{array}{cc}\tan \theta & =\pm \frac{2\sqrt{\frac{a^2(\sqrt{2}+1)}{2}}\sqrt{\frac{a^2(\sqrt{2}-1)}{2}}}{{\left(\sqrt{\frac{a^2(\sqrt{2}-1)}{2}}\right)}^2-{\left(\sqrt{\frac{a^2(\sqrt{2}+1)}{2}}\right)}^2}\\ \tan \theta & =\pm \frac{a^2}{\frac{a^2(\sqrt{2}-1-\sqrt{2}-1)}{2}}\\ & =\pm \frac{a^2}{-a^2}\\ \tan \theta & =\pm -1\\ \tan \theta & =\pm 1\\ \theta & =\pi /4\text{and}3\pi /4\end{array}$$

• y=x3$y=x^3$ and 6y=7−x2$6y=7-x^2$

The two curves meet at point (1,1)$(1,1)$.

For first curve,

dydx=3x2$$\begin{array}{cc}\frac{dy}{dx}& =3x^2\end{array}$$

For second curve,

dydx=−x3$$\begin{array}{cc}\frac{dy}{dx}& =-\frac{x}{3}\end{array}$$

At point (1,1)$(1,1)$, slope of first curve is 3$3$ and second curve is −13$-\frac{1}{3}$.

The product of two slopes is −1$-1$. So angle between them is π/2$\pi /2$.

• x2−y2=2a2$x^2-y^2=2a^2$ and x2+y2=4a2$x^2+y^2=4a^2$

Plotting of these curves reveals this graph:

The two curves intersect at point x=√3a$x=\sqrt{3}a$ and y=a$y=a$.

In our case, for first curve:

dydx=−fxfy=−2x−2y=xy$$\begin{array}{cc}\frac{dy}{dx}& =-\frac{f_x}{f_y}\\ & =-\frac{2x}{-2y}\\ & =\frac{x}{y}\end{array}$$

For second curve,

dydx=−fxfy=−2x2y=−xy$$\begin{array}{cc}\frac{dy}{dx}& =-\frac{f_x}{f_y}\\ & =-\frac{2x}{2y}\\ & =-\frac{x}{y}\end{array}$$

So angle between the two curves is,

tanθ=±m1−m21+m1m2=±xy+xy1−xyxy=±2xyy2−x2$$\begin{array}{cc}\tan \theta & =\pm \frac{m_1-m_2}{1+m_1{m}_2}\\ & =\pm \frac{\frac{x}{y}+\frac{x}{y}}{1-\frac{x}{y}\frac{x}{y}}\\ & =\pm \frac{2xy}{y^2-x^2}\end{array}$$

Substituting for values with point of intersection:

tanθ=±2×√3×aa2−3a2tanθ=±√3θ= π/3 and 2π/3$$\begin{array}{cc}\tan \theta & =\pm \frac{2\times \sqrt{3}\times a}{a^2-3a^2}\\ \tan \theta & =\pm \sqrt{3}\\ \theta & =\pi /3\text{and}2\pi /3\end{array}$$

4.2.2 Find the condition for the curves ax2+by2=1$a{x}^2+b{y}^2=1$ and a1x2+b1y2=1$a_1{x}^2+b_1{y}^2=1$ to intersect orthogonally.

Lets assume the two curves intersect orthogonally at point (h,k)$(h,k)$.

At point (h,k)$(h,k)$, the equation of first curve is ah2+bk2=1$a{h}^2+b{k}^2=1$ and the equation of second curve is a1h2+b1k2=1$a_1{h}^2+b_1{k}^2=1$. Subtracting one from other,

(a−a1)h2=(b1−b)k2h2k2=b1−ba−a1$$\begin{array}{}\begin{array}{cc}(a-a_1)h^2& =(b_1-b)k^2\\ \frac{h^2}{k^2}& =\frac{b_1-b}{a-a_1}\end{array}& \text{(4.1)}\end{array}$$

Slope of first curve,

dydx=−fxfy=−2ax2by=−axby$$\begin{array}{cc}\frac{dy}{dx}& =-\frac{f_x}{f_y}\\ & =-\frac{2ax}{2by}\\ & =-\frac{ax}{by}\end{array}$$

Slope of second curve,

dydx=−fxfy=−2a1x2b1y=−a1xb1y$$\begin{array}{cc}\frac{dy}{dx}& =-\frac{f_x}{f_y}\\ & =-\frac{2a_{1}x}{2b_{1}y}\\ & =-\frac{a_{1}x}{b_{1}y}\end{array}$$

dydx$\frac{dy}{dx}$ of first curve at point (h,k)$(h,k)$ is −ahbk$-\frac{ah}{bk}$ and dydx$\frac{dy}{dx}$ of second curve at point (h,k)$(h,k)$ is −a1hb1k$-\frac{a_{1}h}{b_{1}k}$.

For the two curves to be orthogonal at (h,k)$(h,k)$, product of slopes of two curves at this point should be −1$-1$. i.e.

aa1h2bb1k2=−1$$\begin{array}{}\begin{array}{cc}\frac{a{a}_1{h}^2}{b{b}_1{k}^2}& =-1\end{array}& \text{(4.2)}\end{array}$$

From (4.1) and (4.2),

aa1(b1−b)bb1(a−a1)=−1aa1(b1−b)=bb1(a1−a)aa1b1−aa1b=a1bb1−abb1$$\begin{array}{cc}\frac{a{a}_1(b_1-b)}{b{b}_1(a-a_1)}& =-1\\ a{a}_1(b_1-b)& =b{b}_1(a_1-a)\\ a{a}_1{b}_1-a{a}_{1}b& =a_{1}b{b}_1-ab{b}_1\end{array}$$

Dividing both sides by aa1bb1$a{a}_{1}b{b}_1$,

1b−1b1=1a−1a1$$\begin{array}{cc}\frac{1}{b}-\frac{1}{b_1}& =\frac{1}{a}-\frac{1}{a_1}\end{array}$$

4.3 Question 13

Find the angle of intersection of the following pair of curves:

• r=sinθ+cosθ$r=\sin \theta +\cos \theta$ and r=2sinθ$r=2\sin \theta$

For the point of intersection,

sinθ+cosθ=2sinθsinθ=cosθtanθ=1$$\begin{array}{cc}\sin \theta +\cos \theta & =2\sin \theta \\ \sin \theta & =\cos \theta \\ \tan \theta & =1\end{array}$$

The point of intersection is at (√2,π/4)$(\sqrt{2},\pi /4)$.

For first curve, r=sinθ+cosθ$r=\sin \theta +\cos \theta$, taking log both sides and differentiating w.r.t θ$\theta$,

logr=log(sinθ+cosθ)1rdrdθ=cosθ−sinθcosθ+sinθrdθdr=cosθ+sinθcosθ−sinθ=1+tanθ1−tanθ=tanπ/4+tanθ1−tanπ/4×tanθ=tan(π4+θ)tanϕ1=tan(π4+θ)ϕ1=π/4+θ$$\begin{array}{cc}\log r& =\log (\sin \theta +\cos \theta )\\ \frac{1}{r}\frac{dr}{d\theta }& =\frac{\cos \theta -\sin \theta }{\cos \theta +\sin \theta }\\ r\frac{d\theta }{dr}& =\frac{\cos \theta +\sin \theta }{\cos \theta -\sin \theta }\\ & =\frac{1+\tan \theta }{1-\tan \theta }\\ & =\frac{\tan \pi /4+\tan \theta }{1-\tan \pi /4\times \tan \theta }\\ & =\tan (\frac{\pi }{4}+\theta )\\ \tan {\varphi }_1& =\tan (\frac{\pi }{4}+\theta )\\ {\varphi }_1& =\pi /4+\theta \end{array}$$

For second curve, r=2sinθ$r=2\sin \theta$, taking log both sides and differentiating w.r.t θ$\theta$,

logr=log2+logsinθ1rdrdθ=0+cosθsinθrdθdr=tanθtanϕ2=tanθϕ2=θ$$\begin{array}{cc}\log r& =\log 2+\log \sin \theta \\ \frac{1}{r}\frac{dr}{d\theta }& =0+\frac{\cos \theta }{\sin \theta }\\ r\frac{d\theta }{dr}& =\tan \theta \\ \tan {\varphi }_2& =\tan \theta \\ {\varphi }_2& =\theta \end{array}$$

Thus angle between curves is ϕ1−ϕ2${\varphi }_1-{\varphi }_2$.

ϕ1−ϕ2=π/4+θ−θ=π/4$$\begin{array}{cc}{\varphi }_1-{\varphi }_2& =\pi /4+\theta -\theta =\pi /4\end{array}$$

• r2=a2cos2θ$r^2=a^2\cos 2\theta$ and r2=b2sin2θ$r^2=b^2\sin 2\theta$

Equating two curves, we get.

a2cos2θ=b2sin2θtan2θ=a2b2$$\begin{array}{cc}{a}^2\cos 2\theta & =b^2\sin 2\theta \\ \tan 2\theta & =\frac{a^2}{b^2}\end{array}$$

For first curve, taking log both sides,

2logr=loga2+logcos2θ21rdrdθ=0−2sin2θcos2θ=−tan2θrdθdr=−cot2θtanϕ1=−b2a2$$\begin{array}{cc}2\log r& =\log a^2+\log \cos 2\theta \\ 2\frac{1}{r}\frac{dr}{d\theta }& =0-2\frac{\sin 2\theta }{\cos 2\theta }\\ & =-\tan 2\theta \\ r\frac{d\theta }{dr}& =-\cot 2\theta \\ \tan {\varphi }_1& =-\frac{b^2}{a^2}\end{array}$$

For second curve, taking log both sides,

2logr=logb2+logsin2θ21rdrdθ=0+2cos2θsin2θ=cot2θrdθdr=tan2θtanϕ2=a2b2$$\begin{array}{cc}2\log r& =\log b^2+\log \sin 2\theta \\ 2\frac{1}{r}\frac{dr}{d\theta }& =0+2\frac{\cos 2\theta }{\sin 2\theta }\\ & =\cot 2\theta \\ r\frac{d\theta }{dr}& =\tan 2\theta \\ \tan {\varphi }_2& =\frac{a^2}{b^2}\end{array}$$

Since the product of slopes of two curves is −1$-1$, angle of intersection between them is π/2$\pi /2$.

• r=a(1+cosθ)$r=a(1+\cos \theta )$ and r=b(1−cosθ)$r=b(1-\cos \theta )$

For first curve, taking log both sides and differentiating w.r.t θ$\theta$,

logr=loga+log(1+cosθ)1rdrdθ=0−sinθ1+cosθ=−tanθ/2rdθdr=−cotθ/2tanϕ1=tan(π/2+θ/2)ϕ1=π/2+θ/2$$\begin{array}{cc}\log r& =\log a+\log (1+\cos \theta )\\ \frac{1}{r}\frac{dr}{d\theta }& =0-\frac{\sin \theta }{1+\cos \theta }\\ & =-\tan \theta /2\\ r\frac{d\theta }{dr}& =-\cot \theta /2\\ \tan {\varphi }_1& =\tan (\pi /2+\theta /2)\\ {\varphi }_1& =\pi /2+\theta /2\end{array}$$

For second curve, taking log both sides and differentiating w.r.t θ$\theta$,

logr=logb+log(1−cosθ)1rdrdθ=0+sinθ1−cosθ=cotθ/2rdθdr=tanθ/2tanϕ2=tanθ/2ϕ2=θ/2$$\begin{array}{cc}\log r& =\log b+\log (1-\cos \theta )\\ \frac{1}{r}\frac{dr}{d\theta }& =0+\frac{\sin \theta }{1-\cos \theta }\\ & =\cot \theta /2\\ r\frac{d\theta }{dr}& =\tan \theta /2\\ \tan {\varphi }_2& =\tan \theta /2\\ {\varphi }_2& =\theta /2\end{array}$$

Thus angle between curves is ϕ1−ϕ2${\varphi }_1-{\varphi }_2$ i.e π/2$\pi /2$.

• rn=ancosnθ$r^n=a^n\cos n\theta$ and rn=ansinnθ$r^n=a^n\sin n\theta$

For first curve, taking log both sides and differentiating w.r.t θ$\theta$,

nlogr=nloga+logcosnθn1rdrdθ=0−nsinnθcosnθrdθdr=−cosnθsinnθ=−cotnθtanϕ1=tan(π/2+nθ)ϕ1=π/2+nθ$$\begin{array}{cc}n\log r& =n\log a+\log \cos n\theta \\ n\frac{1}{r}\frac{dr}{d\theta }& =0-n\frac{\sin n\theta }{\cos n\theta }\\ r\frac{d\theta }{dr}& =-\frac{\cos n\theta }{\sin n\theta }\\ & =-\cot n\theta \\ \tan {\varphi }_1& =\tan (\pi /2+n\theta )\\ {\varphi }_1& =\pi /2+n\theta \end{array}$$

For second curve, taking log both sides and differentiating w.r.t θ$\theta$,

nlogr=nloga+logsinnθn1rdrdθ=0+ncosnθsinnθrdθdr=sinnθcosnθtanϕ2=tannθϕ2=nθ$$\begin{array}{cc}n\log r& =n\log a+\log \sin n\theta \\ n\frac{1}{r}\frac{dr}{d\theta }& =0+n\frac{\cos n\theta }{\sin n\theta }\\ r\frac{d\theta }{dr}& =\frac{\sin n\theta }{\cos n\theta }\\ \tan {\varphi }_2& =\tan n\theta \\ {\varphi }_2& =n\theta \end{array}$$

Thus angle between curves is ϕ1−ϕ2${\varphi }_1-{\varphi }_2$ i.e π/2+nθ−nθ=π/2$\pi /2+n\theta -n\theta =\pi /2$.

• rn=ancosnθ$r^n=a^n\cos n\theta$ and rn=bnsinnθ$r^n=b^n\sin n\theta$

For first curve, taking log both sides and differentiating w.r.t θ$\theta$,

nlogr=nloga+logcosnθn1rdrdθ=0−nsinnθcosnθrdθdr=−cosnθsinnθ=−cotnθtanϕ1=tan(π/2+nθ)ϕ1=π/2+nθ$$\begin{array}{cc}n\log r& =n\log a+\log \cos n\theta \\ n\frac{1}{r}\frac{dr}{d\theta }& =0-n\frac{\sin n\theta }{\cos n\theta }\\ r\frac{d\theta }{dr}& =-\frac{\cos n\theta }{\sin n\theta }\\ & =-\cot n\theta \\ \tan {\varphi }_1& =\tan (\pi /2+n\theta )\\ {\varphi }_1& =\pi /2+n\theta \end{array}$$

For second curve, taking log both sides and differentiating w.r.t θ$\theta$,

nlogr=nlogb+logsinnθn1rdrdθ=0+ncosnθsinnθrdθdr=sinnθcosnθtanϕ2=tannθϕ2=nθ$$\begin{array}{cc}n\log r& =n\log b+\log \sin n\theta \\ n\frac{1}{r}\frac{dr}{d\theta }& =0+n\frac{\cos n\theta }{\sin n\theta }\\ r\frac{d\theta }{dr}& =\frac{\sin n\theta }{\cos n\theta }\\ \tan {\varphi }_2& =\tan n\theta \\ {\varphi }_2& =n\theta \end{array}$$

Thus angle between curves is ϕ1−ϕ2${\varphi }_1-{\varphi }_2$ i.e π/2+nθ−nθ=π/2$\pi /2+n\theta -n\theta =\pi /2$.

4.4 Question 14

• Show that at any point of the parabola y2=4ax$y^2=4ax$, the subnormal is constant and the subtangent varies as the abscissa on the point of contact.

dydx=2ay$$\begin{array}{cc}\frac{dy}{dx}& =\frac{2a}{y}\end{array}$$

The subnormal is given by yy′$y{y}^{\prime }$.

Subnormal =yy′=y×2ay=2a, which is a constant.$$\begin{array}{cc}\text{Subnormal}& =y{y}^{\prime }\\ & =y\times \frac{2a}{y}\\ & =2a,\text{which is a constant.}\end{array}$$

Similarly,

Subtangent =yy′=y2ay=y22a=4ax2a=2x$$\begin{array}{cc}\text{Subtangent}& =\frac{y}{y^{\prime }}\\ & =\frac{y}{\frac{2a}{y}}\\ & =\frac{y^2}{2a}\\ & =\frac{4ax}{2a}\\ & =2x\end{array}$$

Thus subtangent varies as the abscissa x$x$ on the point of contact.

• Show that at any point of the hyperbola xy=c2$xy=c^2$, the subtangent varies as the abscissa and the subnormal varies as the cube of the ordinate of the point of contact.

For this hyperbola, dydx=−yx$\frac{dy}{dx}=-\frac{y}{x}$.

Subtangent =yy′=y−yx=−x$$\begin{array}{cc}\text{Subtangent}& =\frac{y}{y^{\prime }}\\ & =\frac{y}{-\frac{y}{x}}\\ & =-x\end{array}$$

Thus subtangent varies as the abscissa x$x$.

Subnormal =yy′=y×−yx=−y2x=−y2c2y=−y3c2$$\begin{array}{cc}\text{Subnormal}& =y{y}^{\prime }\\ & =y\times -\frac{y}{x}\\ & =-\frac{y^2}{x}\\ & =-\frac{y^2}{\frac{c^2}{y}}\\ & =-\frac{y^3}{c^2}\end{array}$$

Thus subnormal varies as the cube of the ordinate y$y$ of the point of contact.

• Show that for the curve by2=(x+a)3$b{y}^2=(x+a{)}^3$ the square of the subtangent varies as the subnormal.

Plotting the graph reveals:

The equation can be written as (x+a)3−by2=0$(x+a{)}^3-b{y}^2=0$.

Differentiating w.r.t.x$x$:

dydx=−fxfy=−3(x+a)2−2by=3(x+a)22by$$\begin{array}{cc}\frac{dy}{dx}& =-\frac{f_x}{f_y}\\ & =-\frac{3(x+a{)}^2}{-2by}\\ & =\frac{3(x+a{)}^2}{2by}\end{array}$$

We have to prove (subtangent)2$(\text{subtangent}{)}^2$ ∝$\propto$ subnormal.

Subnormal =yy′=y×3(x+a)22by=3(x+a)22b$$\begin{array}{cc}\text{Subnormal}& =y{y}^{\prime }\\ & =y\times \frac{3(x+a{)}^2}{2by}\\ & =\frac{3(x+a{)}^2}{2b}\end{array}$$

(Subtangent)2=(yy′)2=(y3(x+a)22by)2=(2by23(x+a)2)2=(2(x+a)33(x+a)2)2=4(x+a)29=827b×3(x+a)22b=827b×Subnormal=constant×Subnormal∝ Subnormal$$\begin{array}{cc}(\text{Subtangent}{)}^2& ={\left(\frac{y}{y^{\prime }}\right)}^2\\ & ={\left(\frac{y}{\frac{3(x+a{)}^2}{2by}}\right)}^2\\ & ={\left(\frac{2b{y}^2}{3(x+a{)}^2}\right)}^2\\ & ={\left(\frac{2(x+a{)}^3}{3(x+a{)}^2}\right)}^2\\ & =\frac{4(x+a{)}^2}{9}\\ & =\frac{8}{27}b\times \frac{3(x+a{)}^2}{2b}\\ & =\frac{8}{27}b\times \text{Subnormal}\\ & =\text{constant}\times \text{Subnormal}\\ & \propto \text{Subnormal}\end{array}$$

Hence, for the given curve square of the subtangent varies as the subnormal.

• Show that in any curve: subnormalsubtangent=(length of normallength of tangent)2$$\frac{\text{subnormal}}{\text{subtangent}}={\left(\frac{\text{length of normal}}{\text{length of tangent}}\right)}^2$$

Here,

LHS =subnormalsubtangent=yy′yy′=y′2$$\begin{array}{cc}\text{LHS}& =\frac{\text{subnormal}}{\text{subtangent}}\\ & =\frac{y{y}^{\prime }}{\frac{y}{y^{\prime }}}\\ & =y^{\prime 2}\end{array}$$

RHS=(length of normallength of tangent)2=(y√1+y′2y√(1+1y′2))2=y′2$$\begin{array}{cc}\text{RHS}& ={\left(\frac{\text{length of normal}}{\text{length of tangent}}\right)}^2\\ & ={\left(\frac{y\sqrt{1+y^{\prime 2}}}{y\sqrt{(1+\frac{1}{y^{\prime 2}})}}\right)}^2\\ & =y^{\prime 2}\end{array}$$

Thus LHS =RHS$\text{LHS}=\text{RHS}$.

4.5 Question 15

Show that for the curve r=eθ$r=e^{\theta }$, the polar subtangent is equal to the polar subnormal.

Differentiating w.r.t θ$\theta$,

drdθ=eθ$$\begin{array}{cc}\frac{dr}{d\theta }& =e^{\theta }\end{array}$$

Polar subnormal is given by drdθ$\frac{dr}{d\theta }$ which is eθ$e^{\theta }$.

Polar subtangent is given by r2dθdr$r^2\frac{d\theta }{dr}$ which is

r2dθdr=r2drdθ=eθ×eθeθ=eθ$$\begin{array}{cc}{r}^2\frac{d\theta }{dr}& =\frac{r^2}{\frac{dr}{d\theta }}\\ & =\frac{e^{\theta }\times e^{\theta }}{e^{\theta }}\\ & =e^{\theta }\end{array}$$

Thus polar subtangent is equal to the polar subnormal.

4.6 Question 16

Find the polar subtangent of:

• r=aeθcotα$r=a{e}^{\theta \cot \alpha }$

Taking log both sides,

logr=loga+log(eθcotα)logr=loga+θcotα$$\begin{array}{cc}\log r& =\log a+\log \left(e^{\theta \cot \alpha }\right)\\ \log r& =\log a+\theta \cot \alpha \end{array}$$

Differentiating w.r.t θ$\theta$ both sides,

1rdrdθ=0+cotαrdθdr=tanαtanϕ=tanαor, ϕ=α$$\begin{array}{cc}\frac{1}{r}\frac{dr}{d\theta }& =0+\cot \alpha \\ r\frac{d\theta }{dr}& =\tan \alpha \\ \tan \varphi & =\tan \alpha \\ \text{or,}\varphi & =\alpha \end{array}$$

Polar subtangent is given by r2dθdr$r^2\frac{d\theta }{dr}$ which is rtanϕ$r\tan \varphi$. So subtangent is,

rtanϕ=rtanα$$\begin{array}{cc}r\tan \varphi & =r\tan \alpha \end{array}$$

• r=2a1−cosθ$r=\frac{2a}{1-\cos \theta }$

Taking log both sides,

logr=log2a−log(1−cosθ)1rdrdθ=0−sinθ1−cosθ1rdrdθ=−cotθ2rdθdr=−tanθ2tanϕ=−tanθ2$$\begin{array}{cc}\log r& =\log 2a-\log (1-\cos \theta )\\ \frac{1}{r}\frac{dr}{d\theta }& =0-\frac{\sin \theta }{1-\cos \theta }\\ \frac{1}{r}\frac{dr}{d\theta }& =-\cot \frac{\theta }{2}\\ r\frac{d\theta }{dr}& =-\tan \frac{\theta }{2}\\ \tan \varphi & =-\tan \frac{\theta }{2}\end{array}$$

Polar subtangent is given by r2dθdr$r^2\frac{d\theta }{dr}$ or rtanϕ$r\tan \varphi$.

So subtangent in our case is,

Subtangent=rtanϕ=r×−tanθ2=−2a1−cosθ×sinθ/2cosθ/2=−2a2sin2θ/2×sinθ/2cosθ/2=−2a2sinθ/2cosθ/2=−2asinθ=−2acscθ$$\begin{array}{cc}\text{Subtangent}& =r\tan \varphi \\ & =r\times -\tan \frac{\theta }{2}\\ & =-\frac{2a}{1-\cos \theta }\times \frac{\sin \theta /2}{\cos \theta /2}\\ & =-\frac{2a}{2{\sin }^2\theta /2}\times \frac{\sin \theta /2}{\cos \theta /2}\\ & =-\frac{2a}{2\sin \theta /2\cos \theta /2}\\ & =-\frac{2a}{\sin \theta }\\ & =-2a\csc \theta \end{array}$$

• Show that in an equiangular spiral r=aeθcotα, the tangent is inclined at a constant angle to the radius vector.

Taking log both sides,

logr=loga+log(eθcotα)logr=loga+θcotα

Differentiating w.r.t θ both sides,

1rdrdθ=0+cotαrdθdr=tanαtanϕ=tanαor, ϕ=α

Thus the tangent is inclined at a constant angle α to the radius vector.

4.7 Question 17

Show that the portion of the tangent to xy=c2 included between the co-ordinate axes is bisected at the point of tangency.

We have the curve, xy=c2. Lets say tangent meets the curve at (x1,y1). Differentiating both sides w.r.t x,

dydx=−yx

Equation of tangent to the curve at point (x1,y1), is,

Y−y1=−y1x1(X−x1)Xy1+Yx1=2x1y1

X2x1+Y2y1=1

This is an equation in double intercept form. So this tangent intersects x-axis at point (2x1,0) and y-axis at point (0,2y1).

The mid-point between these two points is,

0+2x12,2y1+02=x1,y1

which is indeed the point of tangency. Hence, the tangent is bisected at (x1,y1) between the co-ordinate axes.

4.8 Question 18

Prove that

tanϕ=xdydx−yx+ydydx

where ϕ is the angle which the tangent to a curve makes with the radius vector drawn from the origin.

Slope of tangent to curve y=f(x) is tanψ=dydx.

Let P(x,y) be any point on the curve and P(r,θ) be any polar point on the same curve. By the relation between cartesian and polar coordinates, x=rcosθ and r=sinθ. Thus from the graph and simple trigonometry, slope of radius vector passing through origin is tanθ=y/x.

We know well that,

ϕ=ψ−θtanϕ=tan(ψ−θ)tanϕ=tanψ−tanθ1+tanψtanθ=dydx−yx1+dydxyx=1x(xdydx−y)1x(x+ydydx)tanϕ=xdydx−yx+ydydx

4.9 Question 19

Find the pedal equation of the following cartesian curves.

1. y2=4a(x+a)
2. x2/3+y2/3=a2/3
3. c2(x2+y2)=x2y2

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Lets solve first one:

y2=4a(x+a)

Differentiating both sides w.r.t x,

y2=4a(x+a)2ydydx=4a+0dydx=2ay

We have to find relation between p and r to find a pedal equation.

p=xdydx−y√1+(dydx)2=2axy−y√1+4a2y2=2ax−y2√y2+4a2=2ax−4ax−4a2√y2+4a2=−2a(x+2a)√4ax+8a2p=−2a(x+2a)√4a(x+2a)p2=4a2(x+2a)24a(x+2a)p2=a(x+2a)

We know x2+y2=r2, so

x2+4ax+4a2=r2(x+2a)2=r2p4a2=r2p2=ar

Therefore, the required pedal equation, viz. a relation involving p (the length of the perpendicular from the pole to the tangent) and r (the radius vector) for this problem is p2=ar.

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Lets solve second one:

x2/3+y2/3=a2/3

The graph is like this:

For this equation,

x=acos3θ,y=asin3θ

So

dydx=dydθdxdθ

dydθ=3asin2θcosθdxdθ=−3asinθcos2θdydx=−3asin2θcosθ3asinθcos2θ=−tanθ

Lets visualise this to enhance understanding.

The equation of tangent is thus:

y−asin3t=−tant(x−acos3t)

Plotting this tangent on above curve reveals this:

We have to find relation between p and r to find a pedal equation.

p=xdydx−y√1+(dydx)2=−xtanθ−y√1+tan2θ=−xsinθcosθ−y√1+sin2θcos2θ=−xsinθ+ycosθ√sin2θ+cos2θ=−xsinθ+ycosθ√1p=−asinθcos3θ−asin3θcosθp=−asinθcosθ(cos2θ+sin2θ)p=−asinθcosθ

We know x2+y2=r2, so

(acos3θ)2+(asin3θ)2=r2a2((cos3θ)2+(sin3θ)2)=r2a2((cos2θ)3+(sin2θ)3)=r2a2((cos2θ+sin2θ)((cos2θ)2−cos2θsin2θ+(sin2θ)2))=r2a2(1)((cos2θ)2+(sin2θ)2−cos2θsin2θ)=r2a2((cos2θ+sin2θ)2−2cos2θsin2θ−cos2θsin2θ)=r2a2(1−3cos2θsin2θ)=r2a2(1−3(sinθcosθ)2)=r2a2(1−3(−pa)2)=r2a2(a2−3p2a2)=r2a2=r2+3p2

Lets solve third one:

c2(x2+y2)=x2y2

The graph is like this:

Differentiating w.r.t x both sides,

2xc2+2yc2dydx=2x2ydydx+2xy2dydx=2xy2−2xc22yc2−2x2ydydx=x(y2−c2)y(c2−x2)

We have to find relation between p and r to find a pedal equation.

p=xdydx−y√1+(dydx)2=xx(y2−c2)y(c2−x2)−y√1+(x(y2−c2)y(c2−x2))2=x2y2−x2c2−y2c2+x2y2√y2(c2−x2)2+x2(y2−c2)2=2x2y2−c2(x2+y2)√y2(c4−2c2x2+x4)+x2(y4−2y2c2+c4)=2x2y2−x2y2√y2c4−2x2y2c2+x4y2+x2y4−2x2y2c2+x2c4=x2y2√c4(x2+y2)+x2y2(x2+y2)−4x2y2c2p2=x4y4c4(x2+y2)+c2(x2+y2)(x2+y2)−4c2(x2+y2)c2=x4y4(x2+y2)(c4−4c4)+c2(x2+y2)21p2=−3c4(x2+y2)+c2(x2+y2)2x4y41p2=−3c4(x2+y2)+c2(x2+y2)2(c2(x2+y2))2=−3x2+y2+1c2

We know x2+y2=r2, so

1p2=−3r2+1c21p2+3r2=1c2

4.10 Question 20

Find the pedal equation of the following polar curves.

• r=a(1+cosθ)

The graph looks like:

Taking log and differentiating w.r.t θ,

logr=loga+log(1+cosθ)1rdrdθ=0−11+cosθsinθrdθdr=−1+cosθsinθrdθdr=−2cos2θ/22sinθ/2cosθ/2rdθdr=−cotθ/2tanϕ=−cotθ/2tanϕ=tan(π/2+θ/2)ϕ=π/2+θ/2

So pedal equation is given by p=rsinϕ,

p=rsin(π/2+θ/2)p=rcosθ/2

From original equation:

r=a(1+cosθ)r=2acos2θ/2cosθ/2=√r2a

Substituting in above equation gives,

p=r×√r2a

Squaring both sides,

p2=r2×r2a

Thus pedal equation is,

r3=2ap2

• r=2a1−cosθ

Taking log and differentiating w.r.t θ,

logr=log2a−log(1−cosθ)1rdrdθ=−11−cosθ×sinθrdθdr=−1−cosθsinθ=−2sin2θ/22sinθ/2cosθ/2tanϕ=−tanθ/2tanϕ=tan(−θ/2)So, ϕ=−θ/2

Pedal equation is given by p=rsinϕ,

p=rsin(−θ/2)p=−rsin(θ/2)

From original equation:

r=2a1−cosθ2sin2θ/2=2a/rsinθ/2=√ar

Substituting in above equation gives,

p=−r√ar

Squaring both sides,

p2=r2×ar

Pedal equation is thus,

p2=ar

• r2cos2θ=a2

Differentiating w,r.t θ both sides,

r2dcos2θdθ+cos2θdr2dθ=0−2r2sin2θ+2rcos2θdrdθ=02r2sin2θ=2rcos2θdrdθrdθdr=cot2θtanϕ=tan(π/2−2θ)ϕ=π/2−2θ

Pedal equation is given by p=rsinϕ,

p=rsin(π/2−2θ)p=rcos2θ

From original equation, cos2θ=a2r2. Thus pedal equation is,

p=r×a2r2pr=a2

• rm=amcosmθ

Taking log and differentiating w.r.t θ,

mlogr=mloga+log(cosmθ)m1rdrdθ=0−1cosmθmsinmθrdθdr=−cotmθtanϕ=−cotmθtanϕ=tan(π/2+mθ)ϕ=π/2+mθ

Pedal equation is given by p=rsinϕ,

p=rsin(π/2+mθ)p=rcosmθp=r×rmampam=rm+1

• r=aeθcotα

The graph looks like:

Taking log and differentiating w.r.t θ,

logr=loga+log(eθcotα)logr=loga+θcotα1rdrdθ=0+cotαrdθdr=tanαtanϕ=tanαϕ=α

Pedal equation is given by p=rsinϕ. So,

p=rsinα