Differentiation (Part-38) | S N De

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

General Solution of Trigonometric Equation

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 

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

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