Differentiation (Part-38) | S N De

 

Tangent and Normal | Part-9

Here, we are going to solve few Short Answer Type Questions from S N Dey Mathematics, Class 12 from " Tangent and Normal " Chapter.

34. Find the equation of normal to the parabola $~y^2=12x~$ at $~(3t^2, 6t)~$. Hence, find the equation of the normal to this parabola which makes an angle $~135^{\circ}~$ with the x-axis. 

Solution.

$y^2=12x \longrightarrow(1)$

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

$2y~\frac{dy}{dx}=12 \Rightarrow \frac{dy}{dx}=\frac{12}{2y}=\frac 6y.$

$\therefore~~ \left[\frac{dy}{dx}\right]_{(3t^2,6t)}=\frac{6}{6t}=\frac 1t$

$ \\ \Rightarrow \left[-\frac{dx}{dy}\right]_{(3t^2,6t)}=-t.$

So, the equation of the normal at $~(3t^2,6t)~$ is 

$y-6t=-t(x-3t^2) \\ \text{or,}~~ y-6t=-tx+3t^3 \\ \text{or,}~~ xt+y=6t+3t^3 \longrightarrow(2)$

If the normal makes an angle $~135^{\circ}~$ with $x-$axis , then 

$-t=\tan 135^{\circ} \\ \text{or,}~~ -t=\tan(90^{\circ}+45^{\circ}) \\ \text{or,}~~ -t=-\cot 45^{\circ} \\ \therefore~ t=1.$

Putting the value of $~t=1~$ in (2), we get 

$x \times 1+y=6 \times 1+3 \times 1^3 \\ \text{or,}~~ x+y=9~~\text{(ans)}$

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35. Prove that the normals at the points $~(1, 2)~$ and $~(4, 4)~$ of the parabola $~y^2 = 4x~$ intersect on the parabola.

Solution.

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

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

$2y~\frac{dy}{dx}=4 \Rightarrow \frac{dy}{dx}=\frac{4}{2y}=\frac 2y$

$\therefore~ \left[-\frac{dx}{dy}\right]_{(1,2)}=\left[-\frac y2\right]_{(1,2)}=-\frac 22=-1$

$\text{Also,}~\left[-\frac{dx}{dy}\right]_{(1,2)}=-\frac 42=-2. $

$\therefore~$ The equation of the normal at $(1,2)$ is 

$y-2=-1(x-1) \Rightarrow x+y=3\rightarrow(2)$

The equation of the normal at $(4,4)$ is 

$y-4=-2(x-4) \Rightarrow 2x+y=12\rightarrow(3)$

From (2) and (3), we get

$x=3-y \\ \therefore~~ 2(3-y)+y=12 \Rightarrow y=-6$

$\therefore~~ x=3-(-6)=9.$

So, the point of intersection of the normals (2)

and (3) is $~(9,-6).$Clearly, the point $~(9,-6)~$ satisfies the parabola

$~y^2=4x.$

Hence, the normals at the points $~(1, 2)~$ and $~(4, 4)~$ of the parabola

$~y^2 = 4x~$ intersect on the parabola.

S.N. De -Tangent and Normal (Part -1)-Class 12

36. Find the length of the normal chord of the parabola $~y^2 = 4x~$

drawn at $~(1, 2)$.

Solution.

$y^2=4x \longrightarrow(1) \\ \therefore~~ 2y~\frac{dy}{dx}=4 \Rightarrow \frac{dy}{dx}=\frac{4}{2y}=\frac 2y.$

$\therefore~~ \left[-\frac{dx}{dy}\right]_{(1,2)}=\left[-\frac y2\right]_{(1,2)}=-\frac 22=-1.$

So, the equation of the normal at $(1,2)$ is 

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

From (1) and (2) we get,

$(3-x)^2=4x \\ \text{or,}~~ 3^2-2 \cdot 3 \cdot x+x^2=4x \\ \text{or,}~~ 9-6x+x^2=4x \\ \text{or,}~~ x^2-10x+9=0 \\ \text{or,}~~ x^2-9x-x+9=0 \\ \text{or,}~~ x(x-9)-1(x-9)=0 \\ \text{or,}~~ (x-9)(x-1)=0 \Rightarrow x=9,1.$

For $x=9,~~ y=3-9=-6$ and for $~x=1,~~y=3-1=2.$

So, the normal at $(1,2)$ intercepts the parabola (1) at the points $~(1,2)~$ and $~(9,-6).$

Hence, the length of the normal chord is 

$\sqrt{(9-1)^2+(-6-2)^2}=\sqrt{8^2+(-8)^2}\\~~=\sqrt{2 \times 64}=8\sqrt{2}~~ \text{unit.}$

37. Find the equation of the normal to the hyperbola $~x^2-y^2=9~$ at the point $~P(5, 4)~$. Prove that the portion of the normal intercepted between the coordinate axes is bisected at $~P.$

Solution.

$x^2-y^2=9 \longrightarrow(1)$

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

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

$\therefore~ \left[-\frac{dx}{dy}\right]_{(5,4)}=\left[-\frac yx\right]_{(5,4)}=-\frac 45.$

The equation of the normal at $(5,4)$ is 

$y-4=-\frac 45(x-5) \\ \text{or,}~~ 5y-20=-4x+20 \\ \text{or,}~~ 4x+5y=40 \rightarrow(2)~~\text{(ans)}$

Equation (2) can be rewritten as 

$\frac{4x}{40}+\frac{5y}{40}=1 \Rightarrow \frac{x}{10}+\frac{y}{8}=1.$

The normal intercepts the x-axis and y-axis at $~A(10,0)~$ and $~B(0,8).$ So, the midpoint of $~AB~$ is $\left(\frac{10+0}{2},\frac{0+8}{2}\right)=(5,4).$

Hence, the point $~P(5,4)~$ is the midpoint of the line $AB$ and thus follows the result.

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38. Find the equation of the normal to the curve $~x^2 = 4y~$, which passes through the point $~(1, 2).$

Solution.

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

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

$~2x=4~\frac{dy}{dx} \Rightarrow \frac{dy}{dx}=\frac{2x}{4}=\frac x2.$

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

The equation of normal at $(2t,t^2)~$ is 

$y-t^2=-\frac 1t(x-2t) \\ \text{or,}~~ yt-t^3=-x+2t \\ \text{or,}~~ x+yt=2t+t^3 \rightarrow(2)$

Since the normal to the given curve (1) passes through the point $(1,2),$

$1+2t=2t+t^3 \Rightarrow t^3=1 \Rightarrow t=1.$

Putting $~t=1~$ in (2), we get

$x+y \times 1=2 \times 1+1^3 \Rightarrow x+y=3~~\text{(ans)}$

39. Find the equations of the normals at the points on the curve

$~y=\frac{x}{1-x^2}~$  where the tangent makes an angle $~45^{\circ}~$ with the  axis of $~x$.

Solution.

$y=\frac{x}{1-x^2} \longrightarrow(1)$

$\therefore~ \frac{dy}{dx}=\frac{(1-x^2) \cdot 1-x(0-2x)}{(1-x^2)^2}\\ \text{or,}~~ \frac{dy}{dx}=\frac{1-x^2+2x^2}{(1-x^2)^2}=\frac{1+x^2}{(1-x^2)^2} \rightarrow(2)$

Let the tangent to the given curve (1) at $(h,k)$ makes an angle $45^{\circ}$ with $x-$axis.

$\therefore~~\left[\frac{dy}{dx}\right]_{(h,k)}=\tan 45^{\circ} \\ \text{or,}~~ \frac{1+h^2}{(1-h^2)^2}=1 \\ \text{or,}~~ h^2+1=h^4-2h^2+1 \\ \text{or,}~~ h^4-3h^2=0 \\ \text{or,}~~ h^2=0,~~h^2-3=0 \\ \therefore~~ h=0, \pm \sqrt{3}.$

Since the point $(h,k)$ lies on the curve (1), $~k=\frac{h}{1-h^2}.$

For $~h=0,~~k=0.$

For $~h=\pm\sqrt{3},~k=\frac{\pm \sqrt{3}}{1-3}=\mp \frac{\sqrt{3}}{2}.$

So, from (2), we get

$\left[\frac{dy}{dx}\right]_{(h,k)}=\frac{1+h^2}{(1-h^2)^2}$

$\text{So,}~~ \left[\frac{dy}{dx}\right]_{(0,0)}=\frac{1+0}{(1-0)^2}=1$

$\therefore~~\left[-\frac{dx}{dy}\right]_{(0,0)}=-1$

Hence, the equation of normal at $(0,0)$ is 

$y-0=-1(x-0) \Rightarrow x+y=0.$

$\text{Again,}~~ \left[\frac{dy}{dx}\right]_{(\pm \sqrt{3}, \mp \sqrt{3}/2)}=\frac{1+3}{(1-3)^2}=\frac 44=1$

$\therefore~~ \left[-\frac{dx}{dy}\right]=-1.$

So, the equation of normal at $( \pm \sqrt{3},\mp \sqrt{3}/2)~$ is 

$y-( \mp \sqrt{3}/2)=-1 \cdot (x- \mp \sqrt{3}) \\ \text{or,}~~ y \pm \frac{\sqrt{3}}{2}=-x \pm \sqrt{3} \\ \text{or,}~~ 2y \pm \sqrt{3}=-2x \pm 2\sqrt{3} \\ \therefore~~ 2(x+y)= \pm \sqrt{3}~~ \text{(ans)}$

40. A tangent is drawn to the curve $~x^2(x-y) + a^2(x+y)=0~$ at the origin. Find the angle it makes with the x-axis.

Solution.

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

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

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

If the tangent makes an angle $~\theta~$ with the $x-$axis, then

$~\tan \theta=-1 =\tan 135^{\circ} \\ \therefore~~ \theta=135^{\circ}~~\text{(ans)}$

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The MTG series is a good resource for NEET, while Arihant is recommended for IIT JEE. When it comes to Chemistry, NCERT is the most important textbook to refer to, with OP Tandon, MS Chouhan, Pradeeps, Morrison and Boyd, Himanshu Pandey, P Bahadur, and SN Sanyal being useful reference books. For Mathematics, NCERT and RD are good starting points, followed by A Das Gupta, MCQ in Maths, and previous year questions from Arihant 39 years. Additional books from Arihant and GRB can also be used if necessary. HC Verma is a good starting point for Physics, followed by direct IIT JEE questions. Other useful books for Physics include Irodov, Resnick, Krotov, as well as those from Arihant and GRB. Focus on practicing previous year IIT JEE questions multiple times for better results, and prioritize self-study while acknowledging the helpfulness of coaching.

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