# Tangent and Normal | Part-9 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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