Chapter 3 Tangents and Normals-I

3.1 Question 1

Find the equations of the tangents and normals to the following curves at the specified points.

$\boldsymbol{x^2 + y^2 = 25}$ at $(3, -4)$

We know,

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

So $\frac{dy}{dx}$ at point $(3,-4)$ is,

\[\begin{equation*} \begin{split} \frac{dy}{dx} &= \frac{3}{4} \end{split} \end{equation*}\]

Thus equation of tangent is $y + 4 = \frac{3}{4}(x-3)$ or $3x - 4y = 25$.

Also the slope of normal at $(3,-4)$ is $-4/3$, so equation of normal is $y +4 = -\frac{4}{3}(x -3 ) \text{ or } 4x + 3y = 0$.

$\boldsymbol{x^3 -3axy + y^3 = 0}$ at $(x_1, y_1)$

\[\begin{equation*} \begin{split} \frac{dy}{dx} &= -\frac{f_x}{f_y}\\ &= -\frac{3x^2-3ay}{3y^2-3ax} = \frac{ay-x^2}{y^2-ax}\\ \end{split} \end{equation*}\]

Thus equation of tangent at $(x_1, y_1)$ is,

\[\begin{equation*} \begin{split} y - y_1 &= \frac{ay_1-x_1^2}{y_1^2-ax_1}(x-x_1)\\ (x_1^2 - ay_1)x + (y_1^2 - ax_1)y &= x_1^3 -ax_1y_1 + y_1^3 - ax_1y_1\\ (x_1^2 - ay_1)x + (y_1^2 - ax_1)y &=ax_1y_1 \end{split} \end{equation*}\]

Now, slope of normal at $(x_1,y_1)$ is $\frac{y_1^2-ax_1}{x_1^2-ay_1}$. So equation of normal at $(x_1,y_1)$ is,

\[\begin{equation*} \begin{split} y-y_1 &= \frac{y_1^2-ax_1}{x_1^2-ay_1}(x-x_1)\\ \end{split} \end{equation*}\]

$\boldsymbol{c^2(x^2+y^2) = x^2y^2}$ at $(c\sec \theta, c \csc \theta)$

We can write the equation as $c^2(x^2+y^2) - x^2y^2 =0$. Differentiating w.r.t $x$,

\[\begin{equation*} \begin{split} \frac{dy}{dx} &= - \frac{f_x}{f_y}\\ &= - \frac{2xc^2-2xy^2}{2yc^2-2yx^2}\\ \frac{dy}{dx}&= - \frac{x(c^2-y^2)}{y(c^2-x^2)}\\ \end{split} \end{equation*}\]

Slope of tangent at $(c\sec\theta, c\csc\theta)$ is,

\[\begin{equation*} \begin{split} &= -\frac{c\sec\theta(c^2-c^2(\csc)^2\theta)}{c\csc\theta(c^2-c^2(\sec)^2\theta)}\\ &= -\frac{\sec\theta(1-(\csc)^2\theta)}{\csc\theta(1-(\sec)^2\theta)}\\ \frac{dy}{dx} &= -\frac{(\cos)^3\theta}{(\sin)^3\theta}\\ \end{split} \end{equation*}\]

Similarly, slope of normal at $(c\sec\theta, c\csc\theta)$ is $\frac{(\sin)^3\theta}{(\cos)^3\theta}$. Thus equation of tangent at $(c\sec\theta, c\csc\theta)$ is,

\[\begin{equation*} \begin{split} y - c\csc\theta &= -\frac{(\cos)^3\theta}{(\sin)^3\theta}(x-c\sec\theta)\\ x(\cos)^3\theta + y(\sin)^3\theta &= c\csc\theta (\sin)^3\theta + c\sec \theta (\cos)^3 \theta\\ x(\cos)^3\theta + y(\sin)^3\theta &= c((\sin)^2\theta + (\cos)^2\theta)\\ x(\cos)^3\theta + y(\sin)^3\theta &= c\\ \end{split} \end{equation*}\]

Equation of normal at $(c\sec\theta, c\csc\theta)$ is,

\[\begin{equation*} \begin{split} y - c\csc\theta &= \frac{(\sin)^3\theta}{(\cos)^3\theta}(x-c\sec\theta)\\ x(\sin)^3\theta - y(\cos)^3\theta &= c((\sin)^3\theta\sec\theta - c\csc\theta(\cos)^3\theta)\\ x(\sin)^3\theta - y(\cos)^3\theta &= \frac{c((\sin)^2\theta-(\cos)^2\theta)}{\sin\theta \cos\theta} \end{split} \end{equation*}\]

$\boldsymbol{x = a(\theta + \sin \theta), y=a(1-\cos\theta)}$ at $\theta=\pi/2$

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

\[\begin{equation*} \begin{split} \frac{dx}{d\theta} &= a(1+\cos\theta)\\ \frac{dy}{d\theta} &= a(0 + \sin \theta) = a\sin\theta\\ \end{split} \end{equation*}\]

We know,

\[\begin{equation*} \begin{split} \frac{dy}{dx} &= \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}\\ &= \frac{a\sin\theta}{a(1+\cos\theta)} = \frac{\sin\theta}{1+\cos\theta}\\ &= \tan(\theta/2)\\ \frac{dy}{dx} &= \tan(\theta/2)\\ \end{split} \end{equation*}\]

Thus slope of tangent at $\theta=\pi/2$ is $\tan(\pi/4)$ or $1$. Also the slope of normal at same point is $-1$. Hence equation of tangent is,

\[\begin{equation*} \begin{split} y-a(1-\cos\theta) = 1(x-a(\theta + \sin \theta))\\ y-a(1-\cos\pi/2) = 1(x-a(\pi/2 + \sin \pi/2))\\ x-y-\frac{\pi a}{2}=0\\ \end{split} \end{equation*}\]

Also equation of normal is,

\[\begin{equation*} \begin{split} y-a(1-\cos\theta) = -1(x-a(\theta + \sin \theta))\\ y-a(1-\cos\pi/2) = -1(x-a(\pi/2 + \sin \pi/2))\\ y - a = -1(x - \frac{\pi a}{2}-a)\\ x + y = \frac{\pi a}{2} + 2a\\ \end{split} \end{equation*}\]

$\boldsymbol{x^{2/3} + y^{2/3} = a^{2/3}}$ at $(h,k)$

\[\begin{equation*} \begin{split} \text{Here,}\\ \frac{dy}{dx} &= -\frac{f_x}{f_y}\\ &= -\left(\frac{y}{x}\right)^{1/3}\\ \end{split} \end{equation*}\]

Slope of tangent at $(h,k)$ is $-\left(\frac{k}{h}\right)^{1/3}$. And, slope of normal at $(h,k)$ is $\left(\frac{h}{k}\right)^{1/3}$. Thus equation of tangent at $(h,k)$ is,

\[\begin{equation*} \begin{split} y - k &= -\left(\frac{k}{h}\right)^{1/3}(x-h)\\ k^{1/3}x + h^{1/3}y &= k^{1/3}h + kh^{1/3}\\ &= h^{1/3}k^{1/3}(h^{2/3}+k^{2/3})\\ k^{1/3}x + h^{1/3}y &= h^{1/3}k^{1/3}a^{2/3}\\ \end{split} \end{equation*}\]

Similarly equation of normal at $(h,k)$ is,

\[\begin{equation*} \begin{split} y-k &=\left(\frac{h}{k}\right)^{1/3}(x-h)\\ h^{1/3}x-k^{1/3}y &= -kk^{1/3} + h^{1/3}h\\ h^{1/3}x-k^{1/3}y &= h^{4/3} - k^{4/3} \end{split} \end{equation*}\]

$\boldsymbol{y(x-2)(x-3)-x+7=0}$ at the point where it crosses the axis of $x$

The curve crosses the $x$-axis when $y=0$, thus the curve crosses the $x$-axis at $(7,0)$.

Now, lets find out the slope of the curve at that point $(7,0)$.

Expansion of the equation gives: $x^2y-5xy-x+6y+7=0$. Differentiating w.r.t $x$,

\[\begin{equation*} \begin{split} \frac{dy}{dx} &= -\frac{f_x}{f_y}\\ &= -\frac{2xy-5y-1}{x^2-5x+6}\\ \end{split} \end{equation*}\]

At point $(7,0)$, the slope of tangent is $\frac{1}{20}$. Similarly, at point $(7,0)$, the slope of normal is $-20$. Thus the equation of tangent is,

\[\begin{equation*} \begin{split} y-0&= \frac{1}{20}(x-7)\\ x-20y-7=0\\ \end{split} \end{equation*}\]

Equation of normal is,

\[\begin{equation*} \begin{split} y-0 &= -20(x-7)\\ 20x+y &= 140\\ \end{split} \end{equation*}\]

3.2 Question 2

Show that the tangents to the points where the curve $\boldsymbol{y=(x-1)(x-2)(x-3)}$ is met by the $x$-axis, two are parallel and the third makes an angle of $135^0$ with the $x$-axis.

From the equation, the curve meets the $x$-axis at three points: $(1,0), (2,0)$ and $(3,0)$. Expansion of the equation gives,

\[\begin{equation*} \begin{split} y &= x^3-6x^2+11x-6\\ \frac{dy}{dx} &= 3x^2-12x+11\\ \end{split} \end{equation*}\]

Slope of the tangent,

\[\begin{equation*} \begin{split} \frac{dy}{dx} @ (1,0) &= 2\\ \frac{dy}{dx} @ (2,0) &= -1\\ \frac{dy}{dx} @ (3,0) &= 2\\ \end{split} \end{equation*}\]

Thus tangents at $(1,0)$ and $(3,0)$ are parallel as they have same slopes and at point $(3,0)$:

$\tan \psi =-1$ and $\psi=135^0$.

3.3 Question 3

Determine the points of the horizontal and vertical tangents for the following curves:

$\boldsymbol{ax^2 + 2hxy + by^2 =1}$

\[\begin{equation*} \begin{split} \frac{dy}{dx} &= -\frac{f_x}{f_y}\\ &= -\left(\frac{2ax+2hy}{2by+2hx}\right)\\ \end{split} \end{equation*}\]

For the horizontal tangent,

\[\begin{equation*} \begin{split} \frac{dy}{dx} &=0\\ -\left(\frac{2ax+2hy}{2by+2hx}\right) &=0\\ 2ax+2hy &=0\\ ax+hy &=0\\ \end{split} \end{equation*}\]

This is an equation of straight line. And the point of horizontal tangent is given by the point of intersection of this line and the given curve.

For the vertical tangent,

\[\begin{equation*} \begin{split} \frac{dx}{dy} &=0\\ -\left(\frac{2by+2hx}{2ax+2hy}\right) &=0\\ hx+by &=0\\ \end{split} \end{equation*}\]

This is an equation of straight line. And the point of vertical tangent is given by the point of intersection of this line and the given curve.

$\boldsymbol{y= (x-3)^2(x-2)}$

Expansion of equation gives,

\[\begin{equation*} \begin{split} y &= x^3 -8x^2 +21x-18\\ \frac{dy}{dx} &= 3x^2 -16x+21\\ \end{split} \end{equation*}\]

For horizontal tangent,

\[\begin{equation*} \begin{split} \frac{dy}{dx} &= 0\\ 3x^2 -16x+21 &= 0\\ (x-3)(3x-7) &=0\\ \end{split} \end{equation*}\]

Thus $x=3$ and $x=\frac{7}{3}$. For $x=3$, $y=0$ and for $x=\frac{7}{3}$, $y=\frac{4}{27}$.

So, points of horizontal tangents are $(3,0)$ and $(\frac{7}{3},\frac{4}{27})$.

For vertical tangents,

\[\begin{equation*} \begin{split} \frac{dx}{dy} &= 0\\ \frac{1}{3x^2 -16x+21} &= 0\\ \end{split} \end{equation*}\]

Thus for normals, no such points exist.

$\boldsymbol{x^3 + y^3 =3axy}$

The equation can be rewritten as:

$x^3-3axy+y^3=0$

Differentiating w.r.t $x$,

\[\begin{equation*} \begin{split} \frac{dy}{dx} &= -\frac{f_x}{f_y}\\ \frac{dy}{dx} &= -\frac{3x^2-3ay}{3y^2-3ax}\\ &= - \frac{x^2-ay}{y^2-ax}\\ \end{split} \end{equation*}\]

For tangent,

\[\begin{equation*} \begin{split} \frac{dy}{dx} &=0\\ \text{or, } x^2&=ay\\ \end{split} \end{equation*}\]

Solving this and original equation $x^3+y^3=3axy$ gives,

\[\begin{equation*} \begin{split} x^3+y^3 &= 3x \times x^2\\ x^3+y^3 &= 3x^3\\ y^3 &= 2x^3\\ \end{split} \end{equation*}\]

For normal,

\[\begin{equation*} \begin{split} \frac{dx}{dy} &=0\\ \text{or, } y^2&=ax\\ \end{split} \end{equation*}\]

Solving this and original equation $x^3+y^3=3axy$ gives,

\[\begin{equation*} \begin{split} x^3+y^3 &= 3y \times y^2\\ x^3+y^3 &= 3y^3\\ 2y^3 &= x^3\\ \end{split} \end{equation*}\]

3.4 Question 4

Find the points on the curve $y=x^2+3x+4$, the tangents at which pass through the origin.

Differentiating w.r.t $x$,

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

Let the tangent meet the curve at point$(h,k)$. Slope at $(h,k)$ is $2h+3$. The equation of tangent passing through $(h,k)$ is

\[\begin{equation*} \begin{split} y-k &= (2h+3)(x-h)\\ \end{split} \end{equation*}\]

This also passes through origin, so:

\[\begin{equation*} \begin{split} 0-k &= -(2h+3)h\\ 2h^2+3h-k&=0\\ 2h^2+3h-h^2-3h-4 &= 0 \text{, from original equation}\\ h^2 &=0 \\ h &= \pm 2\\ \end{split} \end{equation*}\]

The points on the curve at which tangents pass through origin are:

\[\begin{equation*} \begin{split} (2, 4+6+4) & \text{and } (-2, 4-6+4)\\ \text{or, }(2, 14) & \text{and } (-2,2)\\ \end{split} \end{equation*}\]

Find the point on the curve $y=x^4-6x^3+13x^2-10x+5$ where the tangent is parallel to $y=2x$.

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

\[\begin{equation*} \begin{split} \frac{dy}{dx} &= 4x^3-18x^2 +26x-10\\ \end{split} \end{equation*}\]

For a tangent to be parallel to $y = 2x$,

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

So,

\[\begin{equation*} \begin{split} 4x^3-18x^2 +26x-10 &=2\\ 4x^3-18x^2 +26x-12 &= 0\\ \end{split} \end{equation*}\]

Solving the equation,

\[\begin{equation*} \begin{split} (x-1)(x-2)(2x-3) &=0\\ \end{split} \end{equation*}\]

The values of $x$ are thus $x=1, x=2$ and $x=\frac{3}{2}$.

Therefore the tangents to the curve are parallel to $y=2x$ at points $(1,3), (2,5) \text{and } (\frac{3}{2},\frac{65}{16})$.

3.5 Question 5

Tangents are drawn from the origin to the curve $y=\sin x$. Prove that their points of contacts lie on $x^2y^2=x^2-y^2$.

Let the tangent passing through origin meets the curve tangentially at point $(a,b)$. The slope of the curve is,

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

The slope at $(a,b)$ is $\cos a$. The equation of tangent passing through $(a,b)$ is thus,

\[\begin{equation*} \begin{split} y-b &= \cos a (x-a)\\ \end{split} \end{equation*}\]

As this tangent also passes through origin, it can be written as:

\[\begin{equation*} \begin{split} 0-b &= \cos a (0-a)\\ b^2 &= a^2 \cos^2 a \\ b^2 &= a^2 (1-\sin^2 a)\\ b^2 &= a^2(1-b^2)\\ a^2b^2 &= a^2 -b^2\\ \end{split} \end{equation*}\]

Thus this is the equation of tangent which can be generalised as: $x^2y^2 = x^2-y^2$.

3.6 Question 6

If the tangent at $(x_1,y_1)$ to the curve $x^3 + y^3 =a^3$ meets the curve again in $(x_2,y_2)$, show that $\frac{x_2}{x_1}+\frac{y_2}{y_1}=-1$.

Differentiating w.r.t $x$:

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

The slope of tangent at $(x_1,y_1)$ is $-\frac{x_1^2}{y_1^2}$. The curve also passes through point $(x_2,y_2)$.

Slope of line (in our case, tangent) between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $\frac{y_2-y_1}{x_2-x_1}$.

So,

\[\begin{equation} \begin{split} \frac{y_2-y_1}{x_2-x_1} &= -\frac{x_1^2}{y_1^2}\\ \end{split} \tag{3.1} \end{equation}\]

Equation for the curve at $(x_1,y_1)$ is:

\[\begin{equation*} \begin{split} x_1^3 + y_1^3 &= a^3 \end{split} \end{equation*}\]

Equation for the curve at $(x_2,y_2)$ is:

\[\begin{equation*} \begin{split} x_2^3 + y_2^3 &= a^3 \end{split} \end{equation*}\]

Subtracting one from another, we have:

\[\begin{equation*} \begin{split} x_1^3 - x_2^3 &= y_2^3 - y_1^3\\ \frac{y_2^3 - y_1^3}{x_2^3 - x_1^3} &= -1\\ \frac{(y_2-y_1)(y_2^2 + y_1y_2 + y_1^2)}{(x_2-x_1)(x_2^2+x_1x_2+x_1^2)} &=-1\\ \end{split} \end{equation*}\]

\[\begin{equation} \begin{split} \frac{y_2-y_1}{x_2-x_1} &= -\frac{x_2^2+x_1x_2+x_1^2}{y_2^2 + y_1y_2 + y_1^2} \end{split} \tag{3.2} \end{equation}\]

Solving (3.1) and (3.2), we get,

\[\begin{equation*} \begin{split} -\frac{x_1^2}{y_1^2} &= -\frac{x_2^2+x_1x_2+x_1^2}{y_2^2 + y_1y_2 + y_1^2}\\ x_1^2y_2^2 + x_1^2y_1y_2 + x_1^2y_1^2 &= x_2^2y_1^2 + x_1x_2y_1^2 + x_1^2y_1^2\\ x_1^2y_2^2 + x_1^2y_1y_2 &= x_2^2y_1^2 + x_1x_2y_1^2\\ x_1^2y_2^2 - x_2^2y_1^2 &= x_1x_2y_1^2 - x_1^2y_1y_2\\ (x_1y_2 + x_2y_1)(x_1y_2 - x_2y_1) &= x_1y_1(x_2y_1-x_1y_2)\\ x_1y_2 + x_2y_1 &= - x_1y_1\\ \end{split} \end{equation*}\]

Dividing by $x_1y_1$,

\[\begin{equation*} \begin{split} \frac{x_2}{x_1} + \frac{y_2}{y_1} &= -1\\ \end{split} \end{equation*}\]

3.7 Question 7

Show that a normal at any point of the curve $x=a\cos\theta + a \theta\sin \theta$, $y=a\sin \theta - a\theta\cos\theta$ is at a constant distance from the origin.

Differentiating both equations w.r.t $\theta$:

\[\begin{equation*} \begin{split} \frac{dx}{d\theta} &= -a \sin \theta + a(\cos \theta + \sin \theta) = a\theta \cos \theta\\ \frac{dy}{d\theta} &= a\cos\theta - a(-\theta \sin \theta + \cos \theta) = a\theta \sin \theta \end{split} \end{equation*}\]

So,

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

Slope of normal is $-\cot \theta$.

The equation of normal, therefore is:

\[\begin{equation*} \begin{split} y-(a\sin \theta - a\theta\cos\theta) &= -\cot \theta (x - (a\cos\theta + a \theta\sin \theta))\\ y-a\sin \theta + a\theta\cos\theta &= -\frac{\cos\theta}{\sin \theta} (x - a\cos\theta - a \theta\sin \theta)\\ y\sin\theta - a\sin^2\theta + a\theta\cos\theta\sin\theta &= -x\cos\theta + a\cos^2\theta + a\theta\sin\theta\cos\theta\\ y\sin\theta - a\sin^2\theta - a\cos^2\theta + x\cos\theta &= a\theta\sin\theta\cos\theta - a\theta\cos\theta\sin\theta\\ y\sin\theta + x\cos\theta -a &=0\\ x\cos\theta + y\sin\theta - a &=0 \end{split} \end{equation*}\]

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 distance of the normal from origin is:

\[\begin{equation*} \begin{split} p &= \frac{|0+0-a|}{\sqrt{\cos^2\theta+\sin^2\theta}}\\ p &= \frac{a}{\sqrt{1}}\\ p &= a, \text{which is a constant.}\\ \end{split} \end{equation*}\]

Hence proved.

3.8 Question 8

Show that the curve $\left(\frac{x}{a}\right)^n + \left(\frac{y}{b}\right)^n=2$ touches the straight line $\frac{x}{a}+\frac{y}{b}=2$ at the point $(a,b)$ whatever may be the value of $n$.

Differentiating the curve $\left(\frac{x}{a}\right)^n + \left(\frac{y}{b}\right)^n=2$ w.r.t $x$:

\[\begin{equation*} \begin{split} \frac{dy}{dx} &= -\frac{f_x}{f_y}\\ &= -\frac{\frac{nx^{n-1}}{a^n}}{\frac{ny^{n-1}}{b^n}}\\ \frac{dy}{dx} &= -\frac{b^nx^{n-1}}{a^ny^{n-1}} \end{split} \end{equation*}\]

Slope of tangent to the curve at $(a,b)$ is

\[\begin{equation*} \begin{split} \left(\frac{dy}{dx}\right)_{(a,b)} &= -\frac{b^na^{n-1}}{a^nb^{n-1}}\\ &= -\frac{b}{a} \end{split} \end{equation*}\]

Equation of the tangent to curve at $(a,b)$ is:

\[\begin{equation*} \begin{split} y-b &= -\frac{b}{a}(x-a)\\ ay -ab &= ab -bx\\ bx+ay &= 2ab\\ \frac{x}{a} + \frac{y}{b} &=2 \text{, which is independent of $n$} \end{split} \end{equation*}\]

Prove that $\frac{x}{a}+\frac{y}{b}=1$ touches the curve $y=b e^{-x/a}$ at the point where the curve crosses the axis of $y$.

The curve crosses the $y$-axis at $(0,b)$.

Differentiating curve $\frac{x}{a}+\frac{y}{b}=1$ w.r.t $x$:

\[\begin{equation*} \begin{split} \frac{dy}{dx} &= - b e^{-x/a}\times \frac{1}{a}\\ &= -\frac{y}{a} \end{split} \end{equation*}\]

The equation of tangent to the curve at $(0,b)$ is:

\[\begin{equation*} \begin{split} y-b &= -\frac{b}{a}(x-0)\\ bx + ay &= ab\\ \frac{x}{a} + \frac{y}{b} &=1\\ \end{split} \end{equation*}\]

3.9 Question 9

If $x\cos\alpha + y\sin \alpha =p$ touches the curve $\frac{x^m}{a^m}+\frac{y^m}{b^m}=1$, show that $(a\cos \alpha)^{m/(m-1)}+(b\sin \alpha)^{m/(m-1)}=p^{m/(m-1)}$.

Differentiating the curve $\frac{x^m}{a^m}+\frac{y^m}{b^m}=1$ w.r.t $x$:

\[\begin{equation*} \begin{split} \frac{dy}{dx} &= -\frac{\frac{mx^{m-1}}{a^m}}{\frac{my^{m-1}}{b^m}}\\ \frac{dy}{dx} &= -\frac{b^mx^{m-1}}{a^my^{m-1}}\\ \end{split} \end{equation*}\]

Equation of the tangent is:

\[\begin{equation*} \begin{split} Y-y &= -\frac{b^mx^{m-1}}{a^my^{m-1}}(X-x)\\ b^mx^{m-1}X + a^my^{m-1}Y &= a^mb^m\\ \end{split} \end{equation*}\]

$x\cos\alpha + y\sin \alpha =p$ is the tangent to the curve, so

\[\begin{equation*} \begin{split} \frac{\cos \alpha}{b^mx^{m-1}} &= \frac{\sin \alpha}{a^my^{m-1}} = \frac{p}{a^mb^m}\\ \end{split} \end{equation*}\]

Thus,

\[\begin{equation*} \begin{split} pb^mx^{m-1} &= a^mb^m\cos \alpha\\ x &= \left(\frac{a^m\cos \alpha}{p}\right)^{\frac{1}{m-1}}\\ \end{split} \end{equation*}\]

Also,

\[\begin{equation*} \begin{split} pa^my^{m-1} &=a^mb^m\sin \alpha\\ y &= \left(\frac{b^m \sin \alpha}{p}\right)^{\frac{1}{m-1}}\\ \end{split} \end{equation*}\]

Substituting these values in the curve:

\[\begin{equation*} \begin{split} \frac{\left(\frac{a^m\cos \alpha}{p}\right)^{\frac{m}{m-1}}}{a^m} + \frac{\left(\frac{b^m \sin \alpha}{p}\right)^{\frac{m}{m-1}}}{b^m} &= 1\\ a^{\frac{m^2}{m-1}-m}(\cos \alpha)^{\frac{m}{m-1}} + b^{\frac{m^2}{m-1}-m}(\sin \alpha)^{\frac{m}{m-1}} &= p^{m/(m-1)}\\ a^{\frac{m}{m-1}}(\cos \alpha)^{\frac{m}{m-1}} + b^{\frac{m}{m-1}}(\sin \alpha)^{\frac{m}{m-1}} &= p^{m/(m-1)}\\ (a\cos \alpha)^{\frac{m}{m-1}} + (b\sin \alpha)^{\frac{m}{m-1}} &= p^{m/(m-1)}\\ \end{split} \end{equation*}\]

Prove that the condition that $x\cos \alpha + y\sin \alpha =p$ should touch the curve $x^my^n=a^{m+n}$ is $p^{m+n}m^n n^n=(m+n)^{m+n}a^{m+n}\sin^n\alpha\cos^m\alpha$

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

\[\begin{equation*} \begin{split} \frac{dy}{dx} &= -\frac{f_x}{f_y}\\ &= -\frac{mx^{m-1}y^n}{nx^my^{n-1}}\\ \frac{dy}{dx} &= -\frac{my}{nx}\\ \end{split} \end{equation*}\]

The equation of the tangent to the curve is:

\[\begin{equation*} \begin{split} Y-y &= -\frac{my}{nx}(X-x)\\ myX+nxY &= xmy + nxy\\ \frac{mX}{xm+nx} + \frac{nY}{my+ny} &= 1 \end{split} \end{equation*}\]

$x\cos\alpha + y\sin \alpha =p$ is the tangent to the curve, so

\[\begin{equation*} \begin{split} \frac{\cos \alpha}{\frac{m}{xm+nx}} &= \frac{\sin \alpha}{\frac{n}{my+ny}} = p\\ \end{split} \end{equation*}\]

Thus

\[\begin{equation*} \begin{split} (xm+nx)\cos \alpha &= pm\\ x &= \frac{pm}{(m+n)\cos \alpha}\\ \end{split} \end{equation*}\]

Also, similarly

\[\begin{equation*} \begin{split} y &= \frac{pn}{(m+n)\sin \alpha}\\ \end{split} \end{equation*}\]

Substituting $x$ and $y$ values in the curve:

\[\begin{equation*} \begin{split} \left(\frac{pm}{(m+n)\cos \alpha}\right)^m \left(\frac{pn}{(m+n)\sin \alpha}\right)^m &=a^{m+n}\\ p^{m+n}m^n n^n &=(m+n)^{m+n}a^{m+n}\sin^n\alpha\cos^m\alpha\\ \end{split} \end{equation*}\]

If $\alpha$ and $\beta$ be the intercepts on the axes of $x$ and $y$ cut off by the tangent to the curve $\left(\frac{x}{a}\right)^n + \left(\frac{y}{b}\right)^n=1$ then show that \[\left(\frac{a}{\alpha}\right)^{n/(n-1)}+ \left(\frac{b}{\beta}\right)^{n/(n-1)}=1\]

The intercepts created by the tangent at $x$-axis is $(\alpha, 0)$ and $y$-axis is $(0, \beta)$. Its slope will be $\frac{\beta-0}{0-\alpha}=-\frac{\beta}{\alpha}$.

Its equation will be:

\[\begin{equation*} \begin{split} y -0 &= -\frac{\beta}{\alpha}(x-\alpha)\\ x\beta + y\alpha &= \alpha \beta \end{split} \end{equation*}\]

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

\[\begin{equation*} \begin{split} \frac{dy}{dx} &= -\frac{f_x}{f_y}\\ &= -\left(\frac{x}{y}\right)^{n-1}\left(\frac{b}{a}\right)^{n} \end{split} \end{equation*}\]

The equation of the tangent to the curve is:

\[\begin{equation*} \begin{split} Y-y &= -\left(\frac{x}{y}\right)^{n-1}\left(\frac{b}{a}\right)^{n}(X-x)\\ a^ny^{n-1}Y - a^ny^n &= x^nb^n - b^nx^{n-1}X\\ b^nx^{n-1}X + a^ny^{n-1}Y &= a^nb^n \end{split} \end{equation*}\]

As the two tangent equations represent the same tangent:

\[\begin{equation*} \begin{split} \frac{\beta}{b^nx^{n-1}} &= \frac{\alpha}{a^ny^{n-1}} = \frac{\alpha \beta}{a^nb^n}\\ \end{split} \end{equation*}\]

Thus

\[\begin{equation*} \begin{split} \frac{1}{x^{n-1}} &= \frac{\alpha}{a^n}\\ x &= \left(\frac{a^n}{\alpha}\right)^{\frac{1}{n-1}}\\ \end{split} \end{equation*}\]

Similarly,

\[\begin{equation*} \begin{split} \frac{1}{y^{n-1}} &= \frac{\beta}{b^n}\\ y &= \left(\frac{b^n}{\beta}\right)^{\frac{1}{n-1}}\\ \end{split} \end{equation*}\]

Substituting these values in the curve:

\[\begin{equation*} \begin{split} \frac{\left(\frac{a^n}{\alpha}\right)^{\frac{1}{n-1}}}{a^n} + \frac{\left(\frac{b^n}{\beta}\right)^{\frac{1}{n-1}}}{b^n} &=1\\ \frac{a^{n/(n-1)}}{\alpha^{n/(n-1)}} + \frac{b^{n/(n-1)}}{\beta^{n/(n-1)}} &=1\\ \left(\frac{a}{\alpha}\right)^{n/(n-1)} + \left(\frac{b}{\beta}\right)^{n/(n-1)} &=1\\ \end{split} \end{equation*}\]

3.10 Question 10

If $lx+my=1$ is normal to the parabola $y^2=4ax$ prove that $al^3+2alm^2=m^2$.

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

The slope of normal will be $-\frac{y}{2a}$.

The equation of normal is:

\[\begin{equation*} \begin{split} Y-y &= -\frac{y}{2a}(X-x)\\ yX+2aY &= xy + 2ay\\ yX+2aY &= y(x + 2a)\\ \end{split} \end{equation*}\]

This normal and the given normal in question are same, so:

\[\begin{equation*} \begin{split} \frac{l}{y} &= \frac{m}{2a} = \frac{1}{y(x+2a)}\\ x &= \frac{1}{l}-2a\\ y &= \frac{2a}{m(x+2a)} = \frac{2a}{m(\frac{1}{l}-2a+2a)} = \frac{2al}{m}\\ \end{split} \end{equation*}\]

Substituting these values in the curve $y^2=4ax$,

\[\begin{equation*} \begin{split} \left(\frac{2al}{m}\right)^2 &= 4a(\frac{1}{l}-2a)\\ \frac{4a^2l^2}{m^2} &= 4a \times \frac{1-2al}{l}\\ al^3 &= m^2 -2alm^2\\ al^3 + 2alm^2 &=m^2 \end{split} \end{equation*}\]

Find the condition that $x\cos \theta + y\sin \theta =p$ is normal to the curve $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.

For the curve:

\[\begin{equation*} \begin{split} \frac{dy}{dx} &=-\frac{2x/a^2}{2y/b^2}\\ \frac{dy}{dx} &=-\frac{b^2x}{a^2y}\\ \end{split} \end{equation*}\]

The slope for the normal will be $\frac{a^2y}{b^2x}$.

The equation of the normal is:

\[\begin{equation*} \begin{split} Y-y &= \frac{a^2y}{b^2x}(X-x)\\ b^2xY-b^2xy &= a^2yX - a^2xy\\ a^2yX - b^2xY &= a^2xy -b^2xy\\ a^2yX - b^2xY &= xy(a^2 -b^2)\\ \end{split} \end{equation*}\]

The given normal and this normal are same, so:

\[\begin{equation*} \begin{split} \frac{\cos\theta}{a^2y} &= -\frac{\sin \theta}{b^2x} = \frac{p}{xy(a^2-b^2)}\\ \end{split} \end{equation*}\]

So, $x = \frac{pa^2}{\cos\theta(a^2-b^2)}$ and $y = - \frac{pb^2}{\sin\theta(a^2-b^2)}$.

Substituting these values in the given curve:

\[\begin{equation*} \begin{split} \frac{\left(\frac{pa^2}{\cos\theta(a^2-b^2)}\right)^2}{a^2} + \frac{\left(- \frac{pb^2}{\sin\theta(a^2-b^2)}\right)^2}{b^2} &=1\\ \frac{p^2a^4}{\cos^2\theta(a^2-b^2)^2}\times\frac{1}{a^2} + \frac{p^2b^4}{\sin^2\theta(a^2-b^2)^2}\times \frac{1}{b^2} &=1\\ \frac{p^2}{(a^2-b^2)^2}\left(\frac{a^2}{\cos^2\theta} + \frac{b^2}{\sin^\theta}\right) &=1\\ p^2 &= \frac{(a^2-b^2)^2}{a^2\sec^2\theta + b^2\csc^2\theta}\\ p &= \frac{a^2-b^2}{\sqrt{a^2\sec^2\theta + b^2\csc^2\theta}} \end{split} \end{equation*}\]