Chapter 4 Tangents and Normals-II

4.1 Question 11

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

Differentiating w.r.t \(x\):

\[\begin{equation*} \begin{split} \frac{dy}{dx} &= -\frac{f_x}{f_y}\\ \frac{dy}{dx} &= -\frac{\frac{1}{2\sqrt{x}}}{\frac{1}{2\sqrt{y}}}\\ &=-\frac{\sqrt{y}}{\sqrt{x}}\\ \end{split} \end{equation*}\]

The equation of the tangent to the curve is:

\[\begin{equation*} \begin{split} Y-y &=-\frac{\sqrt{y}}{\sqrt{x}}(X-x)\\ \frac{X}{\sqrt{x}} + \frac{Y}{\sqrt{y}} &= \sqrt{x} + \sqrt{y}\\ \frac{X}{\sqrt{x}} + \frac{Y}{\sqrt{y}} &=\sqrt{a}\\ \frac{X}{\sqrt{x}\sqrt{a}} + \frac{Y}{\sqrt{y}\sqrt{a}} &= 1\\ \end{split} \end{equation*}\]

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

\[\begin{equation*} \begin{split} \sqrt{x}\sqrt{a} + \sqrt{y}\sqrt{a} &= \sqrt{a}(\sqrt{x} +\sqrt{y})\\ &= \sqrt{a}\sqrt{a}\\ &= a \text{, which is indeed a constant.} \end{split} \end{equation*}\]

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

For this,

\[\begin{equation*} \begin{split} \frac{dx}{d\theta} &= -3a\cos^2\theta \sin \theta\\ \frac{dy}{d\theta} &= 3a\sin^2\theta\cos\theta\\ \frac{dy}{dx} &=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}\\ &=-\frac{3a\sin^2\theta\cos\theta}{3a\cos^2\theta \sin \theta}\\ \frac{dy}{dx} &= -\tan\theta\\ \end{split} \end{equation*}\]

The equation of tangent is:

\[\begin{equation*} \begin{split} Y-y &= -\tan\theta (X-x)\\ \tan \theta X + Y &= x\tan\theta + y \\ &=a\cos^3 \theta\times \frac{\sin\theta}{\cos\theta} + a\sin^3\theta\\ &= a\sin\theta(\cos^2\theta + \sin^2\theta)\\ \tan \theta X + Y &= a\sin\theta\\ \end{split} \end{equation*}\]

Dividing both sides by \(a\sin\theta\),

\[\begin{equation*} \begin{split} \frac{X}{a\cos\theta} + \frac{Y}{a\sin\theta} &=1 \end{split} \end{equation*}\]

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

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

\[\begin{equation*} \begin{split} (\text{x-intercept})^2 + (\text{y-intercept})^2\\ a^2\cos^2\theta + a^2\sin^2\theta\\ a^2(\cos^2\theta + \sin^2\theta)\\ a^2 \times 1\\ a^2 \end{split} \end{equation*}\]

which is of constant length.

• In a catenary \(y=c \text{ cosh}\frac{x}{c}\), 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 \(\frac{y^2}{c}\).

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 \((x,y)\). 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 \(x\)-axis.

Differentiating w.r.t \(x\):

\[\begin{equation*} \begin{split} \frac{dy}{dx} &= c \text{ sinh}\frac{x}{c} \times \frac{1}{c}\\ &= \text{ sinh}\frac{x}{c} \end{split} \end{equation*}\]

The equation of the tangent is:

\[\begin{equation*} \begin{split} Y-y &= \text{ sinh}\frac{x}{c}(X-x)\\ \text{ sinh}\frac{x}{c} X - Y + \left(y - x\text{ sinh}\frac{x}{c}\right) &= 0\\ \end{split} \end{equation*}\]

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

\[\begin{equation*} \begin{split} \frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}} \end{split} \end{equation*}\]

So in our case the length of ON is thus:

\[\begin{equation*} \begin{split} \text{ON } &= \frac{|\text{ sinh}\frac{x}{c} x +B \times 0 + y -x\text{ sinh}\frac{x}{c}|}{\sqrt{(\text{ sinh}\frac{x}{c})^2+(-1)^2}}\\ &= \frac{y}{\sqrt{(\text{ cosh}\frac{x}{c})^2}}\\ &= \frac{y}{\text{ cosh}\frac{x}{c}}\\ & =\frac{y}{\frac{y}{c}} = c \text{, which is a constant.} \end{split} \end{equation*}\]

Thus the length of the perpendicular from the foot of the ordinate on the tangent is of constant length \(c\).

The length of the normal MP is given by:

\[\begin{equation*} \begin{split} \text{MP } &= y\sqrt{1+y'^2}\\ &= c\text{ cosh}\frac{x}{c}\sqrt{1+(\text{ sinh}\frac{x}{c})^2}\\ &= c \text{ cosh}\frac{x}{c} \text{ cosh}\frac{x}{c}\\ &= c\times \frac{y}{c}\times \frac{y}{c}\\ &= \frac{y^2}{c} \end{split} \end{equation*}\]

Hence proved.

4.2 Question 12

4.2.1 Find the angle between the pair of curves:

• \(\boldsymbol{x^2-y^2 =a^2}\) and \(\boldsymbol{x^2+y^2=\sqrt{2}a^2}\)

The plotting of the graph reveals this curve:

The angle between two curves is:

\[\begin{equation*} \begin{split} \tan \theta &= \pm \frac{m_1-m_2}{1+m_1m_2} \end{split} \end{equation*}\]

where \(m_1\) and \(m_2\) are the slope of two curves.

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

\[\begin{equation*} \begin{split} \frac{dy}{dx} &= -\frac{f_x}{f_y}\\ &= -\frac{2x}{-2y}\\ &= \frac{x}{y}\\ \end{split} \end{equation*}\]

For second curve,

\[\begin{equation*} \begin{split} \frac{dy}{dx} &= -\frac{f_x}{f_y}\\ &= -\frac{2x}{2y}\\ &= -\frac{x}{y}\\ \end{split} \end{equation*}\]

So angle between the two curves is,

\[\begin{equation*} \begin{split} \tan \theta &= \pm \frac{m_1-m_2}{1+m_1m_2}\\ &= \pm \frac{\frac{x}{y} + \frac{x}{y}}{1 - \frac{x}{y}\frac{x}{y}}\\ &= \pm \frac{2xy}{y^2-x^2} \end{split} \end{equation*}\]

Substituting for values with point of intersection:

\[\begin{equation*} \begin{split} \tan \theta &= \pm \frac{2\sqrt{\frac{a^2(\sqrt{2}+1)}{2}}\sqrt{\frac{a^2(\sqrt{2}-1)}{2}}}{\left(\sqrt{\frac{a^2(\sqrt{2}-1)}{2}}\right)^2-\left(\sqrt{\frac{a^2(\sqrt{2}+1)}{2}}\right)^2}\\ \tan \theta &= \pm \frac{a^2}{\frac{a^2(\sqrt{2}-1-\sqrt{2}-1)}{2}}\\ &= \pm \frac{a^2}{-a^2}\\ \tan \theta &= \pm -1\\ \tan \theta &= \pm 1 \\ \theta &= \text{ $\pi/4$ and $3\pi/4$} \end{split} \end{equation*}\]

• \(\boldsymbol{y=x^3}\) and \(\boldsymbol{6y=7-x^2}\)

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

For first curve,

\[\begin{equation*} \begin{split} \frac{dy}{dx} &= 3x^2 \end{split} \end{equation*}\]

For second curve,

\[\begin{equation*} \begin{split} \frac{dy}{dx} &= -\frac{x}{3} \end{split} \end{equation*}\]

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

The product of two slopes is \(-1\). So angle between them is \(\pi/2\).

• \(\boldsymbol{x^2-y^2=2a^2}\) and \(\boldsymbol{x^2+y^2=4a^2}\)

Plotting of these curves reveals this graph:

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

In our case, for first curve:

\[\begin{equation*} \begin{split} \frac{dy}{dx} &= -\frac{f_x}{f_y}\\ &= -\frac{2x}{-2y}\\ &= \frac{x}{y}\\ \end{split} \end{equation*}\]

For second curve,

\[\begin{equation*} \begin{split} \frac{dy}{dx} &= -\frac{f_x}{f_y}\\ &= -\frac{2x}{2y}\\ &= -\frac{x}{y}\\ \end{split} \end{equation*}\]

So angle between the two curves is,

\[\begin{equation*} \begin{split} \tan \theta &= \pm \frac{m_1-m_2}{1+m_1m_2}\\ &= \pm \frac{\frac{x}{y} + \frac{x}{y}}{1 - \frac{x}{y}\frac{x}{y}}\\ &= \pm \frac{2xy}{y^2-x^2} \end{split} \end{equation*}\]

Substituting for values with point of intersection:

\[\begin{equation*} \begin{split} \tan \theta &= \pm \frac{2\times \sqrt{3}\times a}{a^2-3a^2}\\ \tan \theta &= \pm \sqrt{3}\\ \theta &= \text{ $\pi/3$ and $2\pi/3$} \end{split} \end{equation*}\]

4.2.2 Find the condition for the curves \(ax^2+by^2 =1\) and \(a_1x^2+b_1y^2=1\) to intersect orthogonally.

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

At point \((h,k)\), the equation of first curve is \(ah^2 + bk^2 =1\) and the equation of second curve is \(a_1h^2+ b_1k^2 =1\). Subtracting one from other,

\[\begin{equation*} \begin{split} (a-a_1)h^2 &=(b_1-b)k^2\\ \frac{h^2}{k^2} &= \frac{b_1-b}{a-a_1} \end{split} \tag{4.1} \end{equation*}\]

Slope of first curve,

\[\begin{equation*} \begin{split} \frac{dy}{dx} &= -\frac{f_x}{f_y}\\ &= -\frac{2ax}{2by}\\ &= -\frac{ax}{by} \end{split} \end{equation*}\]

Slope of second curve,

\[\begin{equation*} \begin{split} \frac{dy}{dx} &= -\frac{f_x}{f_y}\\ &= -\frac{2a_1x}{2b_1y}\\ &= -\frac{a_1x}{b_1y} \end{split} \end{equation*}\]

\(\frac{dy}{dx}\) of first curve at point \((h,k)\) is \(-\frac{ah}{bk}\) and \(\frac{dy}{dx}\) of second curve at point \((h,k)\) is \(-\frac{a_1h}{b_1k}\).

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

\[\begin{equation*} \begin{split} \frac{aa_1h^2}{bb_1k^2} &= -1\\ \end{split} \tag{4.2} \end{equation*}\]

From (4.1) and (4.2),

\[\begin{equation*} \begin{split} \frac{aa_1(b_1-b)}{bb_1(a-a_1)} &= -1\\ aa_1(b_1-b) &= bb_1(a_1-a)\\ aa_1b_1 - aa_1b &= a_1bb_1 - abb_1\\ \end{split} \end{equation*}\]

Dividing both sides by \(aa_1bb_1\),

\[\begin{equation*} \begin{split} \frac{1}{b} - \frac{1}{b_1} &= \frac{1}{a} - \frac{1}{a_1} \end{split} \end{equation*}\]

4.3 Question 13

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

• \(\boldsymbol{r=\sin\theta+\cos\theta}\) and \(\boldsymbol{r=2\sin\theta}\)

For the point of intersection,

\[\begin{equation*} \begin{split} \sin\theta +\cos\theta &= 2\sin\theta\\ \sin\theta &= \cos\theta\\ \tan \theta &=1 \end{split} \end{equation*}\]

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

For first curve, \(r=\sin\theta+\cos\theta\), taking log both sides and differentiating w.r.t \(\theta\),

\[\begin{equation*} \begin{split} \log r &= \log(\sin\theta +\cos\theta)\\ \frac{1}{r}\frac{dr}{d\theta} &=\frac{\cos\theta-\sin\theta}{\cos\theta+\sin\theta}\\ r\frac{d\theta}{dr} &=\frac{\cos\theta+\sin\theta}{\cos\theta-\sin\theta}\\ &= \frac{1+\tan \theta}{1-\tan \theta}\\ &= \frac{\tan \pi/4 + \tan \theta}{1 - \tan \pi/4 \times \tan \theta}\\ &= \tan(\frac{\pi}{4}+\theta)\\ \tan {\phi}_1 &= \tan(\frac{\pi}{4}+\theta)\\ {\phi}_1 &= \pi/4 + \theta \end{split} \end{equation*}\]

For second curve, \(r=2\sin \theta\), taking log both sides and differentiating w.r.t \(\theta\),

\[\begin{equation*} \begin{split} \log r &= \log 2 + \log \sin \theta\\ \frac{1}{r}\frac{dr}{d\theta} &= 0 + \frac{\cos \theta}{\sin \theta}\\ r \frac{d\theta}{dr} &= \tan \theta\\ \tan {\phi}_2 &= \tan \theta\\ {\phi}_2 &= \theta \end{split} \end{equation*}\]

Thus angle between curves is \({\phi}_1 -{\phi}_2\).

\[\begin{equation*} \begin{split} {\phi}_1 -{\phi}_2 &= \pi/4 + \theta - \theta = \pi/4 \end{split} \end{equation*}\]

• \(\boldsymbol{r^2=a^2\cos2\theta}\) and \(\boldsymbol{r^2 =b^2\sin 2 \theta}\)

Equating two curves, we get.

\[\begin{equation*} \begin{split} a^2 \cos 2\theta &= b^2\sin 2 \theta\\ \tan 2 \theta &= \frac{a^2}{b^2} \end{split} \end{equation*}\]

For first curve, taking log both sides,

\[\begin{equation*} \begin{split} 2\log r &= \log a^2 + \log \cos 2\theta\\ 2\frac{1}{r}\frac{dr}{d\theta} &= 0 -2\frac{\sin 2 \theta}{\cos 2 \theta}\\ &= -\tan 2 \theta\\ r\frac{d\theta}{dr} &= -\cot 2 \theta\\ \tan {\phi}_1 &= - \frac{b^2}{a^2} \end{split} \end{equation*}\]

For second curve, taking log both sides,

\[\begin{equation*} \begin{split} 2\log r &= \log b^2 + \log \sin 2\theta\\ 2\frac{1}{r}\frac{dr}{d\theta} &= 0 + 2\frac{\cos 2 \theta}{\sin 2 \theta}\\ &= \cot 2 \theta\\ r\frac{d\theta}{dr} &= \tan 2 \theta\\ \tan {\phi}_2 &= \frac{a^2}{b^2} \end{split} \end{equation*}\]

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

• \(\boldsymbol{r=a(1+\cos \theta)}\) and \(\boldsymbol{r=b(1-\cos\theta)}\)

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

\[\begin{equation*} \begin{split} \log r &= \log a + \log (1+\cos\theta)\\ \frac{1}{r}\frac{dr}{d\theta} &= 0 -\frac{\sin \theta}{1+\cos\theta}\\ &= -\tan \theta/2\\ r\frac{d\theta}{dr} &= -\cot \theta/2\\ \tan {\phi}_1 &= \tan (\pi/2 + \theta/2)\\ {\phi}_1 &= \pi/2 + \theta/2 \end{split} \end{equation*}\]

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

\[\begin{equation*} \begin{split} \log r &= \log b + \log (1 - \cos\theta)\\ \frac{1}{r}\frac{dr}{d\theta} &= 0 + \frac{\sin \theta}{1-\cos\theta}\\ &= \cot \theta/2\\ r\frac{d\theta}{dr} &= \tan \theta/2\\ \tan {\phi}_2 &= \tan \theta/2\\ {\phi}_2 &= \theta/2 \end{split} \end{equation*}\]

Thus angle between curves is \({\phi}_1 -{\phi}_2\) i.e \(\pi/2\).

• \(\boldsymbol{r^n=a^n\cos n \theta}\) and \(\boldsymbol{r^n=a^n\sin n \theta}\)

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

\[\begin{equation*} \begin{split} n\log r &= n \log a + \log \cos n\theta\\ n \frac{1}{r}\frac{dr}{d\theta} &= 0 - n \frac{\sin n \theta}{\cos n \theta}\\ r\frac{d\theta}{dr} &= -\frac{\cos n \theta}{\sin n \theta}\\ &= -\cot n \theta\\ \tan {\phi}_1 &= \tan(\pi/2 + n\theta)\\ {\phi}_1 &= \pi/2 + n\theta \end{split} \end{equation*}\]

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

\[\begin{equation*} \begin{split} n \log r &= n\log a + \log \sin n\theta\\ n \frac{1}{r}\frac{dr}{d\theta} &= 0 + n\frac{\cos n \theta}{\sin n \theta}\\ r\frac{d\theta}{dr} &= \frac{\sin n \theta}{\cos n \theta}\\ \tan {\phi}_2 &= \tan n \theta\\ {\phi}_2 &= n \theta \end{split} \end{equation*}\]

Thus angle between curves is \({\phi}_1 -{\phi}_2\) i.e \(\pi/2 + n\theta - n\theta = \pi/2\).

• \(\boldsymbol{r^n=a^n\cos n \theta}\) and \(\boldsymbol{r^n=b^n\sin n \theta}\)

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

\[\begin{equation*} \begin{split} n\log r &= n \log a + \log \cos n\theta\\ n \frac{1}{r}\frac{dr}{d\theta} &= 0 - n \frac{\sin n \theta}{\cos n \theta}\\ r\frac{d\theta}{dr} &= -\frac{\cos n \theta}{\sin n \theta}\\ &= -\cot n \theta\\ \tan {\phi}_1 &= \tan(\pi/2 + n\theta)\\ {\phi}_1 &= \pi/2 + n\theta \end{split} \end{equation*}\]

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

\[\begin{equation*} \begin{split} n \log r &= n\log b + \log \sin n\theta\\ n \frac{1}{r}\frac{dr}{d\theta} &= 0 + n\frac{\cos n \theta}{\sin n \theta}\\ r\frac{d\theta}{dr} &= \frac{\sin n \theta}{\cos n \theta}\\ \tan {\phi}_2 &= \tan n \theta\\ {\phi}_2 &= n \theta \end{split} \end{equation*}\]

Thus angle between curves is \({\phi}_1 -{\phi}_2\) i.e \(\pi/2 + n\theta - n\theta = \pi/2\).

4.4 Question 14

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

\[\begin{equation*} \begin{split} \frac{dy}{dx} &= \frac{2a}{y} \end{split} \end{equation*}\]

The subnormal is given by \(yy'\).

\[\begin{equation*} \begin{split} \text{Subnormal } &= yy'\\ &=y\times \frac{2a}{y}\\ &= 2a, \text{ which is a constant.} \end{split} \end{equation*}\]

Similarly,

\[\begin{equation*} \begin{split} \text{Subtangent } &= \frac{y}{y'}\\ &= \frac{y}{\frac{2a}{y}}\\ &= \frac{y^2}{2a}\\ &=\frac{4ax}{2a}\\ &=2x \end{split} \end{equation*}\]

Thus subtangent varies as the abscissa \(x\) on the point of contact.

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

For this hyperbola, \(\frac{dy}{dx}=-\frac{y}{x}\).

\[\begin{equation*} \begin{split} \text{Subtangent } &= \frac{y}{y'}\\ &= \frac{y}{-\frac{y}{x}}\\ &= -x \\ \end{split} \end{equation*}\]

Thus subtangent varies as the abscissa \(x\).

\[\begin{equation*} \begin{split} \text{Subnormal } &= yy'\\ &= y \times -\frac{y}{x}\\ &= -\frac{y^2}{x}\\ &= -\frac{y^2}{\frac{c^2}{y}}\\ &= -\frac{y^3}{c^2} \end{split} \end{equation*}\]

Thus subnormal varies as the cube of the ordinate \(y\) of the point of contact.

• Show that for the curve \(by^2=(x+a)^3\) the square of the subtangent varies as the subnormal.

Plotting the graph reveals:

The equation can be written as \((x+a)^3-by^2 = 0\).

Differentiating w.r.t.\(x\):

\[\begin{equation*} \begin{split} \frac{dy}{dx} &= -\frac{f_x}{f_y}\\ &= -\frac{3(x+a)^2}{-2by}\\ &= \frac{3(x+a)^2}{2by} \end{split} \end{equation*}\]

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

\[\begin{equation*} \begin{split} \text{Subnormal } &= yy'\\ &= y\times \frac{3(x+a)^2}{2by}\\ &= \frac{3(x+a)^2}{2b} \end{split} \end{equation*}\]

\[\begin{equation*} \begin{split} (\text{Subtangent})^2 &= \left(\frac{y}{y'}\right)^2\\ &= \left(\frac{y}{\frac{3(x+a)^2}{2by}}\right)^2\\ &=\left(\frac{2by^2}{3(x+a)^2}\right)^2\\ &=\left(\frac{2(x+a)^3}{3(x+a)^2}\right)^2\\ &= \frac{4(x+a)^2}{9}\\ &= \frac{8}{27}b \times \frac{3(x+a)^2}{2b}\\ &= \frac{8}{27}b \times \text{Subnormal}\\ &= \text{constant} \times \text{Subnormal}\\ & \propto \text{ Subnormal} \end{split} \end{equation*}\]

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

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

Here,

\[\begin{equation*} \begin{split} \text{LHS }&= \frac{\text{subnormal}}{\text{subtangent}}\\ &= \frac{yy'}{\frac{y}{y'}}\\ &= y'^2 \end{split} \end{equation*}\]

\[\begin{equation*} \begin{split} \text{RHS} &= \left(\frac{\text{length of normal}}{\text{length of tangent}}\right)^2\\ &= \left(\frac{y\sqrt{1+y'^2}}{y\sqrt{\left(1+\frac{1}{y'^2}\right)}}\right)^2\\ &= y'^2 \end{split} \end{equation*}\]

Thus \(\text{LHS }=\text{RHS}\).

4.5 Question 15

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

Differentiating w.r.t \(\theta\),

\[\begin{equation*} \begin{split} \frac{dr}{d\theta} &= e^{\theta}\\ \end{split} \end{equation*}\]

Polar subnormal is given by \(\frac{dr}{d\theta}\) which is \(e^{\theta}\).

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

\[\begin{equation*} \begin{split} r^2\frac{d\theta}{dr} &= \frac{r^2}{\frac{dr}{d\theta}}\\ &= \frac{e^{\theta} \times e^{\theta}}{e^{\theta}}\\ &= e^{\theta} \end{split} \end{equation*}\]

Thus polar subtangent is equal to the polar subnormal.

4.6 Question 16

Find the polar subtangent of:

• \(\boldsymbol{r=ae^{\theta \cot \alpha}}\)

Taking log both sides,

\[\begin{equation*} \begin{split} \log r &= \log a + \log (e^{\theta \cot \alpha})\\ \log r &= \log a + \theta \cot \alpha\\ \end{split} \end{equation*}\]

Differentiating w.r.t \(\theta\) both sides,

\[\begin{equation*} \begin{split} \frac{1}{r}\frac{dr}{d\theta} &= 0 + \cot \alpha\\ r \frac{d\theta}{dr} &= \tan \alpha\\ \tan \phi &= \tan \alpha\\ \text{or, } \phi &= \alpha \end{split} \end{equation*}\]

Polar subtangent is given by \(r^2\frac{d\theta}{dr}\) which is \(r\tan \phi\). So subtangent is,

\[\begin{equation*} \begin{split} r\tan\phi &= r\tan \alpha \end{split} \end{equation*}\]

• \(\boldsymbol{r=\frac{2a}{1-\cos\theta}}\)

Taking log both sides,

\[\begin{equation*} \begin{split} \log r &= \log 2a - \log (1-\cos\theta)\\ \frac{1}{r} \frac{dr}{d\theta} &= 0 - \frac{\sin \theta}{1-\cos\theta}\\ \frac{1}{r} \frac{dr}{d\theta} &= -\cot \frac{\theta}{2}\\ r\frac{d\theta}{dr} &= -\tan \frac{\theta}{2}\\ \tan \phi &= -\tan \frac{\theta}{2} \end{split} \end{equation*}\]

Polar subtangent is given by \(r^2\frac{d\theta}{dr}\) or \(r\tan\phi\).

So subtangent in our case is,

\[\begin{equation*} \begin{split} \text{Subtangent} &= r\tan \phi\\ &= r \times -\tan \frac{\theta}{2}\\ &= -\frac{2a}{1-\cos\theta} \times \frac{\sin \theta/2}{\cos \theta/2}\\ &= -\frac{2a}{2\sin^2\theta/2}\times \frac{\sin \theta/2}{\cos\theta/2}\\ &= - \frac{2a}{2\sin\theta/2\cos\theta/2}\\ &= -\frac{2a}{\sin\theta}\\ &= -2a \csc \theta \end{split} \end{equation*}\]

• Show that in an equiangular spiral \(r=ae^{\theta \cot \alpha}\), the tangent is inclined at a constant angle to the radius vector.

Taking log both sides,

\[\begin{equation*} \begin{split} \log r &= \log a + \log (e^{\theta \cot \alpha})\\ \log r &= \log a + \theta \cot \alpha\\ \end{split} \end{equation*}\]

Differentiating w.r.t \(\theta\) both sides,

\[\begin{equation*} \begin{split} \frac{1}{r}\frac{dr}{d\theta} &= 0 + \cot \alpha\\ r \frac{d\theta}{dr} &= \tan \alpha\\ \tan \phi &= \tan \alpha\\ \text{or, } \phi &= \alpha \end{split} \end{equation*}\]

Thus the tangent is inclined at a constant angle \(\alpha\) to the radius vector.

4.7 Question 17

Show that the portion of the tangent to \(xy = c^2\) included between the co-ordinate axes is bisected at the point of tangency.

We have the curve, \(xy = c^2\). Lets say tangent meets the curve at \((x_1, y_1)\). Differentiating both sides w.r.t \(x\),

\[\begin{equation*} \begin{split} \frac{dy}{dx} &= -\frac{y}{x}\\ \end{split} \end{equation*}\]

Equation of tangent to the curve at point \((x_1,y_1)\), is,

\[\begin{equation*} \begin{split} Y - y_1 &= -\frac{y_1}{x_1}(X - x_1)\\ Xy_1 + Yx_1 &= 2x_1y_1\\ \end{split} \end{equation*}\]

\[\begin{equation*} \begin{split} \frac{X}{2x_1} + \frac{Y}{2y_1} &= 1 \end{split} \end{equation*}\]

This is an equation in double intercept form. So this tangent intersects \(x\)-axis at point \((2x_1,0)\) and \(y\)-axis at point \((0, 2y_1)\).

The mid-point between these two points is,

\[\begin{equation*} \begin{split} \frac{0+2x_1}{2}, \frac{2y_1+0}{2} &= x_1, y_1 \end{split} \end{equation*}\]

which is indeed the point of tangency. Hence, the tangent is bisected at \((x_1,y_1)\) between the co-ordinate axes.

4.8 Question 18

Prove that

\[\begin{equation*} \tan \phi = \frac{x\frac{dy}{dx}-y}{x+y\frac{dy}{dx}} \end{equation*}\]

where \(\phi\) 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 \(\tan \psi = \frac{dy}{dx}\).

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

We know well that,

\[\begin{align*} \phi &= \psi - \theta\\ \tan \phi &= \tan(\psi - \theta)\\ \tan \phi &= \frac{\tan \psi - \tan \theta}{1 + \tan \psi\tan \theta}\\ &= \frac{\frac{dy}{dx}-\frac{y}{x}}{1 + \frac{dy}{dx}\frac{y}{x}}\\ &= \frac{\frac{1}{x}(\frac{xdy}{dx}-y)}{\frac{1}{x}(x+y\frac{dy}{dx})}\\ \tan \phi &= \frac{x\frac{dy}{dx}-y}{x+y\frac{dy}{dx}} \end{align*}\]

4.9 Question 19

Find the pedal equation of the following cartesian curves.

1. \(\boldsymbol{y^2 = 4a(x+a)}\)
2. \(\boldsymbol{x^{2/3} + y^{2/3} = a^{2/3}}\)
3. \(\boldsymbol{c^2(x^2 + y^2) = x^2y^2}\)

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Lets solve first one:

\[\begin{equation} y^2 = 4a(x+a) \end{equation}\]

Differentiating both sides w.r.t \(x\),

\[\begin{align*} y^2 &= 4a(x+a) \\ 2y \frac{dy}{dx} &= 4a + 0 \\ \frac{dy}{dx} &= \frac{2a}{y} \end{align*}\]

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

\[\begin{equation*} \begin{split} p & = \frac{x\frac{dy}{dx} - y}{\sqrt{1 + (\frac{dy}{dx})^2}} \\ & = \frac{\frac{2ax}{y} - y}{\sqrt{1 + \frac{4a^2}{y^2}}}\\ & = \frac{2ax - y^2}{\sqrt{y^2 + 4a^2}}\\ & = \frac{2ax-4ax-4a^2}{\sqrt{y^2+4a^2}}\\ & = -\frac{2a(x+2a)}{\sqrt{4ax+8a^2}}\\ p & = -\frac{2a(x+2a)}{\sqrt{4a(x+2a)}}\\ p^2 & = \frac{4a^2(x+2a)^2}{4a(x+2a)}\\ p^2 & = a(x+2a)\\ \end{split} \end{equation*}\]

We know \(x^2+y^2=r^2\), so

\[\begin{align*} x^2 + 4ax + 4a^2 & = r^2 \\ (x+2a)^2 & = r^2\\ \frac{p^4}{a^2} & = r^2\\ p^2 & = ar\\ \end{align*}\]

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 \(p^2 = ar\).

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Lets solve second one:

\[\begin{equation} x^{2/3} + y^{2/3} = a^{2/3} \end{equation}\]

The graph is like this:

For this equation,

\(x = a\cos^3\theta, y = a\sin^3\theta\)

So

\[\begin{equation*} \begin{split} \frac{dy}{dx} & = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} \end{split} \end{equation*}\]

\[\begin{align*} \frac{dy}{d\theta} & = 3a\sin^2\theta \cos\theta\\ \frac{dx}{d\theta} & = - 3a\sin\theta \cos^2\theta\\ \frac{dy}{dx} & = -\frac{3a\sin^2\theta \cos\theta}{3a\sin\theta \cos^2\theta} = -\tan\theta \end{align*}\]

Lets visualise this to enhance understanding.

The equation of tangent is thus:

\[\begin{align*} y-a\sin^3t=-\tan t\left(x-a\cos^3t\right) \end{align*}\]

Plotting this tangent on above curve reveals this:

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

\[\begin{equation*} \begin{split} p & = \frac{x\frac{dy}{dx} - y}{\sqrt{1 + (\frac{dy}{dx})^2}} \\ & = \frac{-x\tan\theta - y}{\sqrt{1 + \tan^2\theta}}\\ & = - \frac{x\frac{\sin\theta}{\cos\theta} - y}{\sqrt{1+\frac{\sin^2\theta}{\cos^2\theta}}}\\ & = - \frac{x\sin\theta + y\cos\theta}{\sqrt{\sin^2\theta+\cos^2\theta}}\\ & = - \frac{x\sin\theta + y\cos\theta}{\sqrt{1}}\\ p & = -a\sin\theta\cos^3\theta - a\sin^3\theta\cos\theta\\ p & = -a\sin\theta\cos\theta(\cos^2\theta + \sin^2\theta)\\ p & = -a\sin\theta\cos\theta \\ \end{split} \end{equation*}\]

We know \(x^2+y^2=r^2\), so

\[\begin{equation*} \begin{split} (a\cos^3\theta)^2 + (a\sin^3\theta)^2 & = r^2\\ a^2((\cos^3\theta)^2 + (\sin^3\theta)^2) & = r^2\\ a^2((\cos^2\theta)^3 + (\sin^2\theta)^3) & = r^2\\ a^2((\cos^2\theta + \sin^2\theta)((\cos^2\theta)^2 - \cos^2\theta \sin^2\theta + (\sin^2\theta)^2)) & = r^2\\ a^2(1)((\cos^2\theta)^2 + (\sin^2\theta)^2 - \cos^2\theta \sin^2\theta) & = r^2\\ a^2((\cos^2\theta + \sin^2\theta)^2 -2\cos^2\theta \sin^2\theta - \cos^2\theta \sin^2\theta) & = r^2\\ a^2(1-3\cos^2\theta \sin^2\theta) & = r^2\\ a^2(1 - 3(\sin\theta\cos\theta)^2) & = r^2\\ a^2(1 - 3\left(-\frac{p}{a}\right)^2) & = r^2\\ a^2\left(\frac{a^2-3p^2}{a^2}\right) & = r^2 \\ a^2 & = r^2 + 3p^2\\ \end{split} \end{equation*}\]

Lets solve third one:

\[\begin{equation*} c^2(x^2 + y^2) = x^2y^2 \end{equation*}\]

The graph is like this:

Differentiating w.r.t \(x\) both sides,

\[\begin{align*} 2xc^2 + 2yc^2\frac{dy}{dx} &= 2x^2y\frac{dy}{dx} + 2xy^2\\ \frac{dy}{dx} &= \frac{2xy^2 - 2xc^2}{2yc^2 - 2x^2y}\\ \frac{dy}{dx} &= \frac{x(y^2-c^2)}{y(c^2 - x^2)} \end{align*}\]

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

\[\begin{equation*} \begin{split} p &= \frac{x\frac{dy}{dx} - y}{\sqrt{1 + (\frac{dy}{dx})^2}} \\ &= \frac{x\frac{x(y^2-c^2)}{y(c^2-x^2)} - y}{\sqrt{1 + (\frac{x(y^2-c^2)}{y(c^2-x^2)})^2} }\\ &=\frac{x^2y^2-x^2c^2-y^2c^2 + x^2y^2}{\sqrt{y^2(c^2-x^2)^2 + x^2(y^2-c^2)^2}}\\ &= \frac{2x^2y^2-c^2(x^2+y^2)}{\sqrt{y^2(c^4-2c^2x^2+x^4) + x^2(y^4-2y^2c^2+c^4)}}\\ &= \frac{2x^2y^2-x^2y^2}{\sqrt{y^2c^4-2x^2y^2c^2+x^4y^2 + x^2y^4-2x^2y^2c^2+x^2c^4}}\\ &= \frac{x^2y^2}{\sqrt{c^4(x^2+y^2) + x^2y^2(x^2+y^2) - 4x^2y^2c^2}}\\ p^2 &= \frac{x^4y^4}{c^4(x^2+y^2) + c^2(x^2+y^2)(x^2+y^2) - 4c^2(x^2+y^2)c^2}\\ &= \frac{x^4y^4}{(x^2+y^2)(c^4-4c^4)+c^2(x^2+y^2)^2}\\ \frac{1}{p^2} &= \frac{-3c^4(x^2+y^2) + c^2(x^2+y^2)^2}{x^4y^4}\\ \frac{1}{p^2} &= \frac{-3c^4(x^2+y^2) + c^2(x^2+y^2)^2}{(c^2(x^2+y^2))^2}\\ &= -\frac{3}{x^2+y^2} + \frac{1}{c^2}\\ \end{split} \end{equation*}\]

We know \(x^2 + y^2 = r^2\), so

\[\begin{equation*} \begin{split} \frac{1}{p^2} &= -\frac{3}{r^2} + \frac{1}{c^2}\\ \frac{1}{p^2} + \frac{3}{r^2} &= \frac{1}{c^2}\\ \end{split} \end{equation*}\]

4.10 Question 20

Find the pedal equation of the following polar curves.

• \(\boldsymbol{r = a(1+\cos\theta)}\)

The graph looks like:

Taking log and differentiating w.r.t \(\theta\),

\[\begin{equation*} \begin{split} \log r &= \log a + \log (1+\cos \theta)\\ \frac{1}{r}\frac{dr}{d\theta} &= 0 - \frac{1}{1+\cos \theta}\sin \theta\\ r\frac{d\theta}{dr} &= -\frac{1+\cos\theta}{\sin \theta}\\ r\frac{d\theta}{dr} &= -\frac{2\cos^2 \theta/2}{2\sin \theta/2 \cos \theta/2}\\ r\frac{d\theta}{dr} &= -\cot \theta/2\\ \tan \phi &= -\cot \theta/2\\ \tan \phi &= \tan (\pi/2 + \theta/2)\\ \phi = \pi/2 + \theta/2\\ \end{split} \end{equation*}\]

So pedal equation is given by \(p = r\sin \phi\),

\[\begin{equation*} \begin{split} p &= r\sin(\pi/2 + \theta/2)\\ p &= r \cos \theta/2 \end{split} \end{equation*}\]

From original equation:

\[\begin{equation*} \begin{split} r &= a(1+\cos\theta)\\ r &= 2a \cos^2 \theta/2\\ \cos\theta/2 &= \sqrt{\frac{r}{2a}}\\ \end{split} \end{equation*}\]

Substituting in above equation gives,

\[\begin{equation*} \begin{split} p &= r \times \sqrt{\frac{r}{2a}}\\ \end{split} \end{equation*}\]

Squaring both sides,

\[\begin{equation*} \begin{split} p^2 &= r^2 \times\frac{r}{2a}\\ \end{split} \end{equation*}\]

Thus pedal equation is,

\[\begin{equation*} \begin{split} r^3 & = 2ap^2 \end{split} \end{equation*}\]

• \(\boldsymbol{r = \frac{2a}{1 - \cos\theta}}\)

Taking log and differentiating w.r.t \(\theta\),

\[\begin{equation*} \begin{split} \log r &= \log 2a - \log(1-\cos \theta)\\ \frac{1}{r}\frac{dr}{d\theta} &= - \frac{1}{1-\cos\theta}\times \sin\theta\\ r\frac{d\theta}{dr} &= -\frac{1-\cos\theta}{\sin\theta}\\ &= -\frac{2\sin^2\theta/2}{2\sin\theta/2\cos\theta/2}\\ \tan \phi &= -\tan\theta/2\\ \tan \phi &= \tan(-\theta/2)\\ \text{So, } \phi &= -\theta/2\\ \end{split} \end{equation*}\]

Pedal equation is given by \(p = r\sin \phi\),

\[\begin{equation*} \begin{split} p &= r\sin(-\theta/2)\\ p &= -r \sin(\theta/2)\\ \end{split} \end{equation*}\]

From original equation:

\[\begin{equation*} \begin{split} r = \frac{2a}{1 - \cos\theta}\\ 2\sin^2\theta/2 &= 2a/r\\ \sin\theta/2 &= \sqrt{\frac{a}{r}}\\ \end{split} \end{equation*}\]

Substituting in above equation gives,

\[\begin{equation*} \begin{split} p &= -r \sqrt{\frac{a}{r}}\\ \end{split} \end{equation*}\]

Squaring both sides,

\[\begin{equation*} \begin{split} p^2 &= r^2 \times\frac{a}{r}\\ \end{split} \end{equation*}\]

Pedal equation is thus,

\[\begin{equation*} \begin{split} p^2 & = ar \end{split} \end{equation*}\]

• \(\boldsymbol{r^2\cos2\theta = a^2}\)

Differentiating w,r.t \(\theta\) both sides,

\[\begin{equation*} \begin{split} r^2\frac{d\cos2\theta}{d\theta} + \cos2\theta\frac{dr^2}{d\theta}&= 0\\ -2r^2\sin2\theta + 2r\cos2\theta\frac{dr}{d\theta} &=0\\ 2r^2\sin2\theta &= 2r\cos2\theta\frac{dr}{d\theta}\\ r\frac{d\theta}{dr} &= \cot 2 \theta\\ \tan \phi &= \tan(\pi/2 - 2 \theta)\\ \phi = \pi/2 - 2 \theta\\ \end{split} \end{equation*}\]

Pedal equation is given by \(p = r\sin \phi\),

\[\begin{equation*} \begin{split} p &= r\sin(\pi/2 - 2 \theta)\\ p &= r\cos 2 \theta\\ \end{split} \end{equation*}\]

From original equation, \(\cos 2 \theta=\frac{a^2}{r^2}\). Thus pedal equation is,

\[\begin{equation*} \begin{split} p &= r \times \frac{a^2}{r^2}\\ pr &=a^2 \end{split} \end{equation*}\]

• \(\boldsymbol{r^m = a^m\cos m \theta}\)

Taking log and differentiating w.r.t \(\theta\),

\[\begin{equation*} \begin{split} m \log r &= m \log a + \log (\cos m \theta)\\ m \frac{1}{r}\frac{dr}{d\theta} &= 0 - \frac{1}{\cos m \theta} m \sin m \theta \\ r\frac{d\theta}{dr} &= -\cot m \theta\\ \tan \phi &= - \cot m \theta\\ \tan \phi &= \tan (\pi/2 + m\theta)\\ \phi = \pi/2 + m\theta \end{split} \end{equation*}\]

Pedal equation is given by \(p = r\sin \phi\),

\[\begin{equation*} \begin{split} p &= r\sin(\pi/2 + m\theta)\\ p &= r \cos m \theta\\ p &= r \times \frac{r^m}{a^m}\\ pa^m &= r^{m+1} \end{split} \end{equation*}\]

• \(\boldsymbol{r = ae^{\theta \cot \alpha}}\)

The graph looks like:

Taking log and differentiating w.r.t \(\theta\),

\[\begin{equation*} \begin{split} \log r &= \log a + \log (e^{\theta \cot \alpha})\\ \log r &= \log a + \theta \cot \alpha\\ \frac{1}{r}\frac{dr}{d\theta} &= 0 + \cot \alpha\\ r \frac{d\theta}{dr} &= \tan \alpha\\ \tan \phi &= \tan \alpha\\ \phi &= \alpha\\ \end{split} \end{equation*}\]

Pedal equation is given by \(p = r\sin \phi\). So,

\[\begin{equation*} \begin{split} p &= r \sin \alpha \\ \end{split} \end{equation*}\]