Differentiation (Part-38) | S N De

Tangent and Normal  Part-6

In the previous article , we have solved few Short Answer Type Questions. In the following article, we are going to discuss/solve few more  Short Answer Type Questions of S.N.Dey Mathematics-Class 12 of the chapter Tangent and Normal (Ex-14).

Find the equations of the tangent and normal to each of the following curves at the specified points $~[7-11].$ 

7. $~x=a\sec\theta,~y=b\tan\theta~$ at the point $~\theta.$

Solution.

$x=a\sec\theta \Rightarrow \frac{dx}{d\theta}=a\sec\theta\tan\theta$

$y=b\tan\theta \Rightarrow \frac{dy}{d\theta}=b\sec^2\theta$

$\therefore~~ \frac{dy}{dx}=\frac{dy/d\theta}{dx/d\theta}=\frac{b\sec^2\theta}{a\sec\theta\tan\theta}=\frac{b\sec\theta}{a\tan\theta}.$

The equation of tangent to the given curve at $~\theta~$ is 

$y-b\tan\theta=\frac{b\sec\theta}{a\tan\theta}(x-a\sec\theta) \\ \text{or,}~~ ay \tan\theta-ab\tan^2\theta=bx\sec\theta-ab\sec^2\theta \\ \text{or,}~~ ay\tan\theta-bx\sec\theta=-ab(\sec^2\theta-\tan^2\theta) \\ \text{or,}~~ bx\sec\theta-ay\tan\theta=ab \\ \therefore~~ \frac xa \sec\theta-\frac yb \tan\theta=1.$

The equation of normal to the given curve at $~\theta~$ is 

$y-b\tan\theta=\left[-\frac{dx}{dy}\right]_{x=\theta}(x-a\sec\theta) \\ \text{or,}~~ y-b\tan\theta=-\frac{a\tan\theta}{b\sec\theta}(x-a\sec\theta) \\ \text{or,}~~ by\sec\theta-b^2\tan\theta\sec\theta=-ax\tan\theta\\~~~+a^2\sec\theta\tan\theta \\ \text{or,}~~ by \sec\theta+ax\tan\theta=(a^2+b^2)\tan\theta\sec\theta \\ \therefore~~ \frac{ax}{\sec\theta}+\frac{by}{\tan\theta}=a^2+b^2~~\text{(ans)}$

8. $~xy^2=18~$ at the point $~(2,3).$

Solution.

$xy^2=18 \longrightarrow(1)$

Differentiating (1) w.r.t. $~x~$, we get

$1 \cdot y^2+ x \cdot 2y~\frac{dy}{dx}=0 \\ \text{or,}~~ \frac{dy}{dx}=-\frac{y^2}{2xy}=-\frac{y}{2x} \\ \therefore~~\left[\frac{dy}{dx}\right]_{(2,3)}=-\frac{3}{2 \times 2}=-\frac 34$

 $\left[-\frac{dx}{dy}\right]_{(2,3)}=\left[\frac{2x}{y}\right]_{(2,3)}=\frac{2 \times 2}{3}=\frac 43$

The equation of tangent to the curve (1) at $~(2,3)~$ is 

$y-3=\left[\frac{dy}{dx}\right]_{(2,3)}(x-2)\\ \text{or,}~~ y-3=-\frac 34(x-2) \\ \text{or,}~~ 4(y-3)=-3(x-2) \\ \text{or,}~~ 4y-12=-3x+6 \\ \therefore~~ 3x+4y=18~~\text{(ans)}$

The equation of normal to the curve (1) at $~(2,3)~$ is 

$y-3=\left[-\frac{dx}{dy}\right]_{(2,3)}(x-2) \\ \text{or,}~~ y-3=\frac 43(x-2) \\ \text{or,}~~ 3y-9=4x-8 \\ \therefore~~ 4x-3y+1=0~~\text{(ans)}$

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9. $~y(x-2)(x-3)+7=x~$ at the point , where the curve intersects the x-axis.

Solution.

$y(x-2)(x-3)+7=x \\ \text{or,}~~ y(x-2)(x-3)=x-7 \longrightarrow(1) \\ \therefore~~ \log[y(x-2)(x-3)]=\log(x-7) \longrightarrow(2)$

Differentiating (2) w.r.t. $~x~$ , we get

$\frac 1y~\frac{dy}{dx}+\frac{1}{x-2}+\frac{1}{x-3}=\frac{1}{x-7} \\ \therefore~~ \frac{dy}{dx}=y\left(\frac{1}{x-7}-\frac{1}{x-2}-\frac{1}{x-3}\right) \\ \therefore~~\frac{dy}{dx}=\frac{x-7}{(x-2)(x-3)}\left[\frac{1}{x-7}-\frac{1}{x-2}-\frac{1}{x-3}\right] \\ \text{or,}~~\frac{dy}{dx}=\frac{1}{(x-2)(x-3)}-\frac{x-7}{(x-2)^2(x-3)}-\frac{x-7}{(x-2)(x-3)^2}$

Clearly, the given curve (1) intersects the x-axis at $~(7,0).$

$\therefore~~\left[\frac{dy}{dx}\right]_{(7,0)}\\=\frac{1}{(7-2)(7-3)}-0-0=\frac{1}{5 \times 4}=\frac{1}{20}$

So, the equation of tangent to the given curve at $~(7,0)~$ is 

$~y-0=\left[\frac{dy}{dx}\right]_{(7,0)}(x-7) \\ \text{or,}~~ y-0=\frac{1}{20}(x-7) \\ \text{or,}~~ 20y=x-7 \Rightarrow x-20y=7~~\text{(ans)}$

The equation of normal to the given curve at $~(7,0)~$ is 

$y-0=\left[-\frac{dx}{dy}\right]_{(7,0)}(x-7) \\ \text{or,}~~y=-20(x-7) \\ \therefore~~ 20x+y=140~~\text{(ans)}$

10. $~x=a(\theta-\sin\theta),~y=a(1-\cos\theta)~$ at $~\theta=\pi.$

Solution.

$x=a(\theta-\sin\theta) \Rightarrow \frac{dx}{d\theta}=a(1-\cos\theta)$

$y=a(1-\cos\theta) \Rightarrow \frac{dy}{d\theta}=a\sin\theta$

$\therefore~~ \frac{dy}{dx}=\frac{dy/d\theta}{dx/d\theta}=\frac{a\sin\theta}{a(1-\cos\theta)}=\frac{\sin\theta}{1-\cos\theta}.$

$\therefore~~\left[\frac{dy}{dx}\right]_{\theta=\pi}=\frac{\sin\pi}{1-\cos\pi}=0.$

$[x]_{\theta=\pi}=a(\pi-\sin\pi)=a\pi$

$[y]_{\theta=\pi}=a(1-\cos\pi)=a[1-(-1)]=2a$

So, the equation of tangent to the given curve at $~(a\pi,2a)~$ is 

$y-2a=0(x-a\pi) \Rightarrow y=2a~~\text{(ans)}$

Again, the equation of normal to the given curve at $~(a\pi,2a)~$ is

$y-2a=\left[-\frac{dx}{dy}\right]_{\theta=\pi}(x-a\pi) \\ \text{or,}~~ x-a\pi=\left[-\frac{dy}{dx}\right]_{\theta=\pi}(y-2a)=0 \cdot (y-2a) \\ \therefore~~ x=a\pi~~\text{(ans)}$

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11. $~x^{2/3}+y^{2/3}=2~$ at $~(1,1).$

Solution.

$x^{2/3}+y^{2/3}=2 \longrightarrow(1)$

Differentiating (1) w.r.t. $~x~$, we get

$\frac 23 x^{\frac 23-1}+\frac 23 y^{\frac 23-1}~\frac{dy}{dx}=0 \\ \text{or,}~~ \frac 23\left(x^{-\frac 13}+y^{-\frac 13}~\frac{dy}{dx}\right)=0 \\ \text{or,}~~ \frac{dy}{dx}=-\frac{x^{-1/3}}{y^{-1/3}}=-\frac{y^{1/3}}{x^{1/3}}=-\sqrt[3]{\frac yx}$

$\therefore~~\left[\frac{dy}{dx}\right]_{(1,1)}=-\sqrt[3]{\frac 11}=-1$

$\left[-\frac{dx}{dy}\right]_{(1,1)}=1$

So, the equation of tangent to the given curve (1) at $~(1,1)~$ is

$y-1=-1(x-1) \Rightarrow x+y=2~~\text{(ans)}$

Also, the equation of normal to the given curve (1) at $~(1,1)~$ is 

$y-1=1(x-1) \Rightarrow  x-y=0~~\text{(ans)}$

12. Show that the equation of the normal to the hyperbola $~x=a\sec\theta,~y=b\tan\theta~$ at the point $~(a\sec\theta,b\tan\theta)~$ is $~ax\cos\theta+by\cot\theta=a^2+b^2.$

Solution.

$x=a\sec\theta \Rightarrow \frac{dx}{d\theta}=a\sec\theta\tan\theta$

$y=b\tan\theta \Rightarrow \frac{dy}{d\theta}=b\sec^2\theta$

$\therefore~~\frac{dy}{dx}=\frac{dy/d\theta}{dx/d\theta}=\frac{b\sec^2\theta}{a\sec\theta\tan\theta}=\frac{b\sec\theta}{a\tan\theta}$

$\text{So,}~~ -\frac{dx}{dy}=-\frac{a\tan\theta}{b\sec\theta}$

Hence, equation of normal to the given curve at $~(a\sec\theta,b\tan\theta)~$ is 

$y-b\tan\theta=-\frac{a\tan\theta}{b\sec\theta}(x-a\sec\theta) \\ \text{or,}~~ by \sec\theta-b^2\tan\theta\sec\theta=-ax\tan\theta\\~~~+a^2\tan\theta\sec\theta \\ \text{or,}~~ ax \tan\theta+by \sec\theta=(a^2+b^2)\sec\theta\tan\theta \\ \text{or,}~~ \frac{ax}{\sec\theta}+\frac{by}{\tan\theta}=a^2+b^2 \\ \therefore~~ ax\cos\theta+by \cot\theta=a^2+b^2.$

13. Find the equation of the tangent to the circle $~x^2+y^2=a^2~$ at the point $~(a\cos\theta, a\sin\theta).~$ Hence show that the line $~y=x+a\sqrt{2}~$ touches the given circle ; find the coordinates of the point of contact.

Solution.

$x^2+y^2=a^2 \longrightarrow(1)$

Differentiating w.r.t. $~x~$, we get

$2x+2y~\frac{dy}{dx}=0 \Rightarrow \frac{dy}{dx}=-\frac xy.$

$\therefore~~ \left[\frac{dy}{dx}\right]_{(a\cos\theta,a\sin\theta)}=-\frac{a\cos\theta}{a\sin\theta}=-\frac{\cos\theta}{\sin\theta}$

The equation of tangent to the given curve (1) at $~(a\cos\theta,a\sin\theta)~$ is 

$y-a\sin\theta=-\frac{\cos\theta}{\sin\theta}(x-a\cos\theta) \\ \text{or,}~~ y\sin\theta-a\sin^2\theta=-x\cos\theta+a\cos^2\theta \\ \text{or,}~~ x\cos\theta+y\sin\theta=a(\sin^2\theta+\cos^2\theta) \\ \therefore~~ x\cos\theta+y\sin\theta=a^2 \longrightarrow(2)$

Now, slope of the straight line $~y=x+a\sqrt{2}~$ is $~1.$

Again, the slope of tangent to the given curve  at any point is given by $~ -\frac{\cos\theta}{\sin\theta}=-\cot\theta.$

$\therefore~~ -\cot\theta=1 \Rightarrow \theta=\frac{3\pi}{4}.$

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From (2) we get the required equation of tangent at $~\theta=\frac{3\pi}{4}~$ is

$x\cos(3\pi/4)+y\sin(3\pi/4)=a \\ \text{or,}~~ x\left(-\frac{1}{\sqrt{2}}\right)+y\left(\frac{1}{\sqrt{2}}\right)=a \\ \text{or,}~~ \frac{1}{\sqrt{2}}y=a+\frac{x}{\sqrt{2}} \\ \text{or,}~~ y=x+a\sqrt{2}.$

Hence, the straight line $~y=x+a\sqrt{2}~$ touches the given circle.

So, coordinates of required point of contact is given by $~\left(a\cos\frac{3\pi}{4}, a\sin\frac{3\pi}{4}\right)=\left(-\frac{a}{\sqrt{2}}, \frac{a}{\sqrt{2}}\right).$