# Tangent and Normal | Part-8 Tangent and Normal  Part-8

Normal in Geometry:

In geometry, the term "normal" refers to a line or vector that is perpendicular, or at a 90-degree angle, to another line, surface, or shape at a specific point. The normal line is often used in the context of curves, surfaces, or polygons.

For instance, the normal line to a curve at a given point is a line that is perpendicular to the tangent line at that point. It provides information about the direction and orientation of the curve at that particular location.

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

27. Show that the equation of the normal to the ellipse  $~\frac{x^2}{25}+\frac{y^2}{9}=1~$ at $~\left(\frac{5}{\sqrt{2}},\frac{5}{\sqrt{2}}\right)~$ is the line $~5x-3y=8\sqrt{2}$.

Solution. $\frac{x^2}{25}+\frac{y^2}{9}=1 \longrightarrow(1)$ Differentiating (1) w.r.t. $x,$ we get $\frac{2x}{25}+\frac{2y}{9}~\frac{dy}{dx}=0 \\ \text{or,}~~ \frac{dy}{dx}=-\frac{2x/25}{2y/9}=-\frac{9x}{25y}.$ $\therefore~~ \left[\frac{dy}{dx}\right]_{(5/\sqrt{2},3/\sqrt{2})}=-\frac{9}{25} \times \frac{5/\sqrt{2}}{3/\sqrt{2}}=-\frac 35.$ So, the slope of the normal to the given curve (1) at the point $\left(\frac{5}{\sqrt{2}},\frac{3}{\sqrt{2}}\right)$ is $~ \frac 53.$ Hence, the equation of normal at the point $\left(\frac{5}{\sqrt{2}},\frac{3}{\sqrt{2}}\right)$ is $y-\frac{3}{\sqrt{2}}=\frac 53 \left(x-\frac{5}{\sqrt{2}}\right) \\ \text{or,}~~ \frac{3\sqrt{2}y-9}{\sqrt{2}}=\frac{5\sqrt{2}x-25}{\sqrt{2}} \\ \text{or,}~~ 3\sqrt{2}y-9=5\sqrt{2}x-25 \\ \text{or,}~~ \sqrt{2}(3y-5x)=9-25 \\ \text{or,}~~ -\sqrt{2}(5x-3y)=-16 \\ \text{or,}~~ 2(5x-3y)=16\sqrt{2} \\ \therefore~~ 5x-3y=8\sqrt{2}~~\text{(ans)}$

28. Show that the equation of the normal to the hyperbola $~\frac{x^2}{a^2}-\frac{y^2}{b^2}=1~$ at the point $~(a\sqrt{2},b)~$ is $~ax+b\sqrt{2}y=(a^2+b^2)\sqrt{2}.$

Solution.

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

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

$\frac{2x}{a^2}-\frac{2y}{b^2}~\frac{dy}{dx}=0 \\ \text{or,}~~ \frac{dy}{dx}=\frac{-2x/a^2}{-2y/b^2}=\frac{xb^2}{ya^2}.$

$\therefore~~ \left[\frac{dy}{dx}\right]_{(a\sqrt{2},b)}=\frac{a\sqrt{2} \cdot b^2}{b \cdot a^2}=\frac{b\sqrt{2}}{a}.$

So, slope of the normal to the curve (1) at the point $~(a\sqrt{2},b)~$ is  $~ -\frac{a}{b\sqrt{2}}.$

The equation of normal at $~(a\sqrt{2},b)~$ is

$y-b=-\frac{a}{b\sqrt{2}}(x-a\sqrt{2}) \\ \text{or,}~~ yb\sqrt{2}-b^2\sqrt{2}=-ax+a^2\sqrt{2} \\ \text{or,}~~ ax+yb \sqrt{2}=\sqrt{2}(a^2+b^2)~~ \text{(ans)}$

#### Solution of S N De-Probability-class 12 -Beng. Version

29. Find the equation of normal to the hyperbola $~x^2-y^2=16~$ at $~(4\sec\theta,4\tan\theta)~$. Hence, show that the line $~x+\sqrt{2}y=8\sqrt{2}~$ is a normal to this hyperbola. Find the coordinates of the foot of the normal.

Solution.

$x^2-y^2=16 \longrightarrow(1)$

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

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

$\therefore~~\left[\frac{dy}{dx}\right]_{(4\sec\theta,4\tan\theta)}=\frac{4 \sec\theta}{4\tan\theta}=\frac{\sec\theta}{\tan\theta}.$

The slope of the normal to the given curve at $~(4 \sec\theta,4\tan\theta)~$ is

$~\left[-\frac{dx}{dy}\right]=-\frac{\tan\theta}{\sec\theta}.$

So, the equation of normal at $~(4\sec\theta,4\tan\theta)~$ is

$y-4\tan\theta=-\frac{\tan\theta}{\sec\theta}(x-4\sec\theta) \\ \text{or,}~~ y\sec\theta-4\tan\theta \sec\theta=-x\tan\theta+4\sec\theta \tan\theta \\ \text{or,}~~ x\tan\theta+y \sec\theta=8\sec\theta \tan\theta \\ \text{or,}~~ \frac{x}{\sec\theta}+\frac{y}{\tan\theta}=8 \\ \text{or,}~~ x \cos\theta+ y\cot\theta=8 \longrightarrow(2)$

Now, the slope of the given straight line $~x+\sqrt{2}y=8\sqrt{2}~$ is $~ -\frac{1}{\sqrt{2}}.$

Since the slope of the normal to the given hyperbola is $~ -\frac{\tan\theta}{\sec\theta},$

$\therefore~~ -\frac{\tan\theta}{\sec\theta}=-\frac{1}{\sqrt{2}} \\ \text{or,}~~ \frac{\sin\theta}{\cos\theta \sec\theta}=\frac{1}{\sqrt{2}} \\ \text{or,}~~ \sin\theta=\frac{1}{\sqrt{2}}=\sin(\pi/4) \\ \therefore~~ \theta=\frac{\pi}{4}.$

Hence, the equation of the normal at $~ \theta=\frac{\pi}{4}~$ is

$x \cdot \cos(\pi/4)+y \cot(\pi/4)=8 \\ \text{or,}~~x \cdot \frac{1}{\sqrt{2}}+y \cdot 1=8 \\ \therefore~~ x+y\sqrt{2}=8\sqrt{2}.$

So, the coordinates of the foot of the normal is $~\left(4\sec \frac{\pi}{4}, 4\tan \frac{\pi}{4}\right)=(4\sqrt{2},4) ~~(\text{ans.})$

30. Find the equation of normal to the parabola $~y^2=4x~$, parallel to the straight line $~y=2x.$

Solution.

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

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

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

Suppose that the normal to the parabola (1) at $(h,k)$ is parallel to the straight line $~y=2x.$

$\therefore~~ -\left(\frac{dx}{dy}\right)_{(h,k)}=2 \\ \text{or,}~~ -\frac k2=2 \\ \text{or,}~~ k=-4.$

Since the point $~(h,k)~$ lies on the given parabola,

$k^2=4h \\ \text{or,}~~ 4h=(-4)^2=16 \\ \text{or,}~~ h=\frac{16}{4}=4.$

So, the normal to the parabola having slope $2$ passes through the point $(4,-4).$

Hence the equation of the normal is given by

$y-(-4)=2(x-4) \\ \text{or,}~~ y+4=2x-8 \\ \text{or,}~~ y=2x-8-4 \\ \therefore~~ y=2x-12~~(\text{ans.})$

31. Find the equation of normal to the ellipse $~x^2+4y^2 = 4~$ at $~(2 \cos\theta, \sin\theta)~$. Hence, find the equation of normal to this ellipse which is parallel to the line $~8x+3y=0.$

Solution.

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

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

$2x+8y~\frac{dy}{dx}=0 \\ \text{or,}~~ \frac{dy}{dx}=-\frac{2x}{8y}=-\frac{x}{4y}.$

$\therefore~~\left[\frac{dy}{dx}\right]_{(2 \cos\theta, \sin\theta)}=-\frac{2 \cos\theta}{4s}=-\frac{\cos\theta}{2\sin\theta}.$

So, $-\left[\frac{dx}{dy}\right]_{(2\cos\theta, \sin\theta)}=\frac{2\sin\theta}{\cos\theta}.$

So, the equation of the normal to the curve at $(2 \cos\theta, \sin\theta)$ is

$y-\sin\theta=-\left[\frac{dx}{dy}\right]_{(2\cos\theta,\sin\theta)}(x-2\cos\theta) \\ \text{or,}~~ y-\sin\theta=\frac{2\sin\theta}{\cos\theta}(x-2\cos\theta) \\ \text{or,}~~ y\cos\theta-\sin\theta \cos\theta=2x \cos\theta-4\sin\theta \cos\theta \\ \text{or,}~~ 2x \sin\theta-y \cos\theta=3 \sin\theta \cos\theta$

Since the normal to the given curve at $(h,k)$ is parallel to the straight line $~8x+3y=0,$ so

$\left[-\frac{dx}{dy}\right]_{(h,k)}=-\frac 83~~[*] \\ \text{or,}~~ \frac{4k}{h}=-\frac 83 \\ \text{or,}~~ h=-4k \times \frac 38=-\frac{3k}{2} \rightarrow(2)$

Note[*] : $8x+3y=0 \Rightarrow~ y=-\frac 83x.$

Since the point $(h,k)$ lies on the ellipse, so

$h^2+4k^2=4 \\ \text{or,}~~ \left(-\frac{3k}{2}\right)^2+4k^2=4 \\ \text{or,}~~ \frac{9k^2}{4}+4k^2=4 \\ \text{or,}~~ \frac{9k^2+16k^2}{4}=4 \\ \text{or,}~~ 25k^2=16 \\ \text{or,}~~ k=\pm \sqrt{\frac{16}{25}}= \pm \frac 45.$

So, from (2) we get,

$h=-\frac 32 \cdot k=-\frac 32 \times \left(\pm \frac 45\right) \\ \text{or,}~~ h= \mp \frac 65.$

So, the equation of the normal at $\left(-\frac 65, \frac 45\right)$ is

$y-\frac 45=-\frac 83 \left[x-(-\frac 65)\right] \\ \text{or,}~~ y-\frac 45=-\frac 83(x+6/5) \\ \text{or,}~~ \frac{5y-4}{5}=-\frac 83x-\frac{16}{5} \\ \text{or,}~~ \frac{5y-4}{5}=\frac{-40x-48}{15} \\ \text{or,}~~ 3(5y-4)=-40x-48 \\ \text{or,}~~ 15y-12=-40x-48 \\ \text{or,}~~ 40x+15y=-36.$

Again, the equation of the normal at $\left(\frac 65,-\frac 45\right)~$ is given by $y-(-4/5)=-\frac 83(x-6/5) \\ \text{or,}~~ y+\frac 45=-\frac 83x+\frac{16}{5} \\ \text{or,}~~ 15 \left(y+\frac 45\right)=15 \left(-\frac 83x+\frac{16}{5}\right) \\ \text{or,}~~ 15y+12=-40x+48 \\ \therefore~ 40x+15y=36.$

#### Chhaya math solution of GENERAL SOLUTIONS OF TRIGONOMETRIC EQUATIONS, class XI

32. Find the equation of that normal to the parabola $~x^2 = 4ay~$ which makes an angle $~60^{\circ}~$ with the x-axis.

Solution.

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

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

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

$\therefore~~\frac{dx}{dy}=\frac{2a}{x}.$

Let the parabola makes an angle $~60^{\circ}~$ with the $x-$ axis at the point $(h,k).$

$\therefore~~ \left[-\frac{dx}{dy}\right]_{(h,k)}=\tan 60^{\circ} \\ \text{or,}~~ -\frac{2a}{h}=\sqrt{3} \Rightarrow h=-\frac{2a}{\sqrt{3}} \rightarrow(2)$

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

$h^2=4ak \\ \text{or,}~~ \left(-\frac{2a}{\sqrt{3}}\right)^2=4ak~~[\text{By (2)}] \\ \text{or,}~~ \frac{4a^2}{3}=4ak \\ \therefore~~ k=\frac a3.$

Now, the equation of the normal at the point $~\left(-\frac{2a}{\sqrt{3}},\frac a3\right)$ is

$y-\frac a3=\sqrt{3}\left[x-\left(-\frac{2a}{\sqrt{3}}\right)\right] \\ \text{or,}~~ y-\frac a3=\sqrt{3} x+2a \\ \text{or,}~~ 3y-a=3\sqrt{3} x+6a \\ \text{or,}~~ 3y=3\sqrt{3}x+7a~~\text{(ans)}$

33. Find the equations of normals to the hyperbola $~3x^2-2y^2 = 10~$ at points where the line $~x+y+3=0~$ cuts the curve.

Solution.

$3x^2-2y^2=10 \longrightarrow(1),~ x+y+3=0 \longrightarrow(2)$

From (2), we get $~ y=-(x+3) \rightarrow(3)$

So, from (1) and (3), we get

$3x^2-2[-(x+3)]^2=10 \\ \text{or,}~~ 3x^2-2(x+3)^2=10 \\ \text{or,}~~ 3x^2-2(x^2+6x+9)=10 \\ \text{or,}~~ x^2-12x-28=0 \\ \text{or,}~~ x^2-14x+2x-28=0 \\ \text{or,}~~ x(x-14)+2(x-14)=0 \\ \text{or,}~~ (x-14)(x+2)=0 \\ \therefore~~ x-14=0,~x+2=0 \\ \text{or,}~~ x=14,-2.$

For $x=14,~~y=-(14+3)=-17;$

For $x=-2,~~ y=-(-2+3)=-1.$

So, the straight line cuts the hyperbola (1) at the points $(14,-17)$ and $(-2,-1).$

Now, $~3x^2-2y^2=10 \\ \therefore~ 6x-4y~\frac{dy}{dx}=0 \\ \text{or,}~~ \frac{dy}{dx}=\frac{6x}{4y}=\frac{3x}{2y}.$

$\therefore~ \left[\frac{dy}{dx}\right]_{(14,-17)}=\frac{3 \times 14}{2 \times (-17)}=-\frac{21}{17}.$

$\text{So,}~~ \left[-\frac{dx}{dy}\right]_{(14,-17)}=\frac{17}{21}.$

The equation of the normal at $(14,-17)$ is

$y+17=\frac{17}{21}(x-14) \\ \text{or,}~~ 21y+357=17x-238 \\ \text{or,}~~ -17x+21y=-238-357 \\ \therefore~~ 17x-21y=595~~\text{(ans)}$

Again, $\left[\frac{dy}{dx}\right]_{(-2,-1)}=\frac{3 \times (-2)}{2 \times (-1)}=3$

$\therefore~~ \left[-\frac{dx}{dy}\right]_{(-2,-1)}=-\frac 13.$

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

$y+1=-\frac 13(x+2) \\ \text{or,}~~ 3y+3=-x-2 \\ \text{or,}~~ x+3y+5=0~~\text{(ans)}$