Differentiation (Part-38) | S N De

Tangent and Normal | Part-5

 

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

1(i) Find the points on the hyperbola $~2x^2-3y^2=6~$ at which the slope of the tangent line is $~(-1).$

Solution.

$2x^2-3y^2=6 \longrightarrow(1)$

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

$4x-6y~\frac{dy}{dx}=0 \Rightarrow \frac{dy}{dx}=\frac{4x}{6y}=\frac{2x}{3y}\rightarrow(2).$

Let $~(h,k)~$ be the point on the hyperbola (1) at which the slope of the tangent line is $~(-1).$

$\therefore~~ \left[\frac{dy}{dx}\right]_{(h,k)}=-1 \\ \text{or,}~~ \frac{2h}{3k}=-1~~[\text{By (2)}] \\ \text{or,}~~ h=-\frac 32k.$

Since the point $~(h,k)~$ lies on the hyperbola (1),

$~2h^2-3k^2=6 \\ \text{or,}~~ 2 \left(-\frac 32k\right)^2-3k^2=6 \\ \text{or,}~~ 2 \times \frac{9k^2}{4}-3k^2=6 \\ \text{or,}~~ \frac{9k^2}{2}-3k^2=6 \\ \text{or,}~~ 9k^2-6k^2=12 \\ \text{or,}~~ 3k^2=12 \\ \text{or,}~~ k^2=\frac{12}{3} \\ \text{or,}~~k=\pm \sqrt{4}=\pm 2.$

For $~k=2,~~ h=-\frac 32 \times 2=-3.$

For $~k=-2,~~ h=-\frac 32 \times (-2)=3.$

Hence, the required points are $~(-3,2)~$ and $~(3,-2).$

(ii) The slope of the tangent line to the curve $~x^3-x^2-2x+y-4=0~$ at some point on it is $~1.~$ Find the coordinates of such point or points.

Solution.

$x^3-x^2-2x+y-4=0 \\ \text{or,}~~ y=4+2x+x^2-x^3 \longrightarrow(1)$

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

$\frac{dy}{dx}=0+2+2x-3x^2=2+2x-3x^2\rightarrow(2)$

Let the slope of the tangent line to the curve (1) at the point $~(h,k)~$ be $~1.$

$\therefore~~ \left[\frac{dy}{dx}\right]_{(h,k)}=1 \\ \text{or,}~~ 2+2h-3h^2=1~~[\text{By (2)}]\\ \text{or,}~~  0=3h^2-2h+1-2 \\ \text{or,}~~  3h^2-2h-1=0 \\ \therefore~~ h=\frac{-(-2) \pm \sqrt{(-2)^2-4 \cdot 3 \cdot (-1)}}{2 \times 3} =\frac{2 \pm \sqrt{4+12}}{6} \\ \text{or,}~~  h=\frac{2 \pm \sqrt{16}}{6}=\frac{2 \pm 4}{6}$

 $\therefore~~h=\frac{2+4}{6}=1,~~ h=\frac{2-4}{6}=-\frac 13.$

Since $~(h,k)~$ lies on (1), $~k=4+2h+h^2-h^3.$

For $~h=1,~~ k=4+2 \times 1+1^2-1^3=6.$

For $~h=-\frac 13,$

$ k=4 + 2 \times \left(-\frac 13\right)+\left(-\frac 13\right)^2-\left(-\frac 13\right)^3 \\ \text{or,}~~ k=4-\frac 23+\frac 19+\frac{1}{27}=4+\frac{-18+3+1}{27} \\ \therefore~~k=3+\left(1-\frac{14}{27}\right)=3+\frac{13}{27}=3\frac{13}{27}.$

Hence, the required points are $~(1,6)~$ and $~\left(-\frac 13,3 \frac{13}{27}\right).$

(iii) The slope of the normal to the parabola $~3y^2+4y+2=x~$ at a point on it is $~8.~$ Find the coordinates of the point.

Solution.

$3y^2+4y+2=x \longrightarrow(1)$

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

$\frac{d}{dx}(3y^2+4y+2)=\frac{d}{dx}(x) \\ \text{or,}~~ 6y~\frac{dy}{dx}+4~\frac{dy}{dx}+0=1 \\ \text{or,}~~ (6y+4)~\frac{dy}{dx}=1 \\ \text{or,}~~ \frac{dy}{dx}=\frac{1}{6y+4} \\ \text{or,}~~ \frac{dx}{dy}=6y+4 \\ \text{or,}~~ -\frac{dx}{dy}=-6y-4$

Let $~(h,k)~$ be the point on the parabola (1) at which the slope of normal to the given curve (1) is $~8.$

$\left[-\frac{dx}{dy}\right]_{(h,k)}=8 \\ \text{or,}~~ -6k-4=8 \\ \text{or,}~~ -6k=8+4 \\ \therefore~~ k=\frac{12}{-6}=-2.$

Since the point $~(h,k)~$ lies on the parabola (1), so  $~3k^2+4k+2=h.$

For $~k=-2,$

$~h=3(-2)^2+4 \times (-2)+2=12-8+2=6.$

Hence, the coordinates of the required point is $~(6,-2).$

2. Find a point on the parabola $~y=x^2-6x+9~,~$ where the tangent is parallel to the line joining the points $~(4,1)~$ and $~(3,0).$

Solution.

$y=x^2-6x+9 \Rightarrow \frac{dy}{dx}=2x-6.$

Let the required point be $~(h,k).$

The slope of the straight line joining the points $~(4,1)~$  and $~(3,0)~$ is $~\frac{1-0}{4-3}=1.$

By question,

$\left[\frac{dy}{dx}\right]_{(h,k)}=1 \\ \text{or,}~~ 2h-6=1 \Rightarrow h=\frac 72.$

Since the point $~(h,k)~$ lies on the given parabola, so $k=h^2-6h+9.$

For $~h=\frac 72,$

$ k=\left(\frac 72\right)^2-6 \left(\frac 72\right)+9=\frac{49}{4}-21+9=\frac 14.$

Hence, the required point is $~\left(\frac 72, \frac 14\right).$

Find the equations of the tangent and normal to each of the following curves at the specified points $~[3-11].$

3. $~y=x^2+4x+1~$ at $~x=3.$

Solution.

$y=x^2+4x+1 \longrightarrow(1)\\ \text{or,}~~ \frac{dy}{dx}=2x+4  \\ \therefore~~ \left[\frac{dy}{dx}\right]_{x=3}=2 \times 3+4=10.$

For $~x=3,~~y=3^2+4 \times 3+1=9+12+1=22.$

So, the equation of tangent to the curve (1) at $~(3,22)~$ is 

$y-22=10(x-3)\Rightarrow 10x-y=8~~\text{(ans)}$

Again, the equation of normal to the curve (1) at $~(3,22)~$ is 

$y-22=\left[-\frac{dx}{dy}\right]_{x=3}(x-3) \\ \text{or,}~~ y-22=\left(\frac{-1}{2x+4}\right)_{x=3}(x-3) \\ \text{or,}~~ y-22=\frac{-1}{2 \times 3+4}(x-3) \\ \text{or,}~~ y-22=\frac{-1}{10}(x-3) \\ \text{or,}~~ 10(y-22)=-x+3 \\ \therefore~~ x+10y=223~~\text{(ans)}$

4. $~y^2=4ax~$ at the ends of latus rectum.

Solution.

The coordinates of the end of the latus rectum of the parabola $~y^2=4ax \rightarrow(1)~$ is $~(a, \pm 2a).$

From (1) we get, $~2y~\frac{dy}{dx}=4a \Rightarrow \frac{dy}{dx}=\frac{4a}{2y}=\frac{2a}{y}.$

$~\left[\frac{dy}{dx}\right]_{(a,2a)}=\frac{2a}{2a}=1.$

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

$~y-2a=1 \cdot (x-a) \Rightarrow x-y+a=0.$

Again, $~\left[\frac{dy}{dx}\right]_{(a,-2a)}=\frac{2a}{-2a}=-1.$

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

$~y-(-2a)=-1 \cdot (x-a) \Rightarrow x+y+a=0.$

2nd Part :

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

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

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

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

5. $~y=x^3-3x~$ at the point $~(2,2).$

Solution.

$y=x^3-3x \longrightarrow(1) \\ \therefore~~ \frac{dy}{dx}=3x^2-3 \\ \text{So,}~~ \left[\frac{dy}{dx}\right]_{(2,2)}=3 \times 2^2-3=12-3=9$

$\therefore~~ \left[-\frac{dx}{dy}\right]_{(2,2)}=\left(\frac{-1}{3x^2-3}\right)_{(2,2)}=-\frac 19.$

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

$y-2=9(x-2) \Rightarrow 9x-y=16~~\text{(ans)}$

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

$y-2=-\frac 19(x-2) \\ \text{or,}~~9(y-2)=-(x-2) \\ \therefore~~ x+9y=20~~ \text{(ans)}$

6. $~\frac{x^2}{a^2}+\frac{y^2}{b^2}=1~$ at $~(a\cos\theta,b \sin\theta)$

Solution.

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \\ \text{or,}~~ b^2x^2+a^2y^2=a^2b^2 \longrightarrow(1)$

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

$~b^2 \cdot 2x+a^2 \cdot 2y~\frac{dy}{dx}=0 \\ \text{or,}~~2\left(b^2x+a^2y~\frac{dy}{dx}\right)=0 \\ \text{or,}~~ \frac{dy}{dx}=-\frac{b^2x}{a^2y} \\ \therefore~~ \left[\frac{dy}{dx}\right]_{(a\cos\theta, b\sin\theta)}=-\frac{b^2 \cdot a \cos\theta}{a^2 \cdot b\sin\theta}=-\frac ba \cdot \frac{\cos\theta}{\sin\theta}$

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

$y-b\sin\theta=-\frac ba \cdot \frac{\cos\theta}{\sin\theta}(x-a\cos\theta) \\ \text{or,}~~ ay \sin\theta-ab\sin^2\theta=-bx\cos\theta+ab\cos^2\theta \\ \text{or,}~~ bx\cos\theta+ay\sin\theta=ab(\cos^2\theta+\sin^2\theta) \\ \text{or,}~~ bx \cos\theta+ay\sin\theta=ab \\ \text{or,}~~ \frac{x\cos\theta}{a}+\frac{y\sin\theta}{b}=1~~\text{(ans)}$

Again, $~-\frac{dx}{dy}=\frac{a^2y}{b^2x} \\ \therefore~~ \left[-\frac{dx}{dy}\right]_{(a\cos\theta, b\sin\theta)}=\frac{a^2}{b^2} \cdot \frac{b\sin\theta}{a\cos\theta}=\frac ab \cdot \frac{\sin\theta}{\cos\theta}.$

So, the equation of normal to the curve (1) at $~(a\cos\theta,b\sin\theta)~$ is 

$y-b\sin\theta=\frac{a}{b} \cdot \frac{\sin\theta}{\cos\theta}(x-a\cos\theta) \\ \text{or,}~~ by \cos\theta-b^2\sin\theta\cos\theta=ax \sin\theta-\\~~a^2\sin\theta\cos\theta \\ \text{or,}~~ by \cos\theta-ax\sin\theta=(b^2-a^2)\sin\theta\cos\theta \\ \text{or,}~~ \frac{by}{\sin\theta}-\frac{ax}{\cos\theta}=b^2-a^2 \\ \therefore~~ \frac{ax}{\cos\theta}-\frac{by}{\sin\theta}=a^2-b^2~~\text{(ans)}$