Differentiation (Part-38) | S N De

Tangent and Normal VSA type Solutions

 

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).

Find the equation of the tangent at the specified points to each of the following curves $~[5-15] :$

5. the parabola $~y^2=4x~$ at $~(1,2).$

Solution.

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

Differentiating both sides of (1) with respect to $~x~$ , we get

$~2y~\frac{dy}{dx}=4 \Rightarrow \frac{dy}{dx}=\frac{4}{2y}=\frac 2y \\ \therefore~~ \left[\frac{dy}{dx}\right]_{(1,2)}=\frac 22=1.$

$\therefore~$ The equation of tangent at $~(1,2)~$ to the parabola $~y^2=4x~$ is given by 

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

6. the ellipse $~9x^2+16y^2=288~$ at $~(4,3)$

Solution.

$~9x^2+16y^2=288 \longrightarrow(1)$

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

$~9 \cdot 2x+16 \cdot 2y~\frac{dy}{dx}=0 \\ \text{or,}~~ 9x+16y~\frac{dy}{dx}=0 \\ \text{or,}~~ \frac{dy}{dx}=-\frac{9x}{16y} \\ \therefore~~ \left[\frac{dy}{dx}\right]_{(4,3)}=-\frac{9 \times 4}{16 \times 3}=-\frac 34.$

$\therefore~~$ The equation of tangent at $~(4,3)~$ to the parabola (1) is given by 

$~y-3= \left[\frac{dy}{dx}\right]_{(4,3)}(x-4) \\ \text{or,}~~ y-3=-\frac 34(x-4) \\ \text{or,}~~ 4(y-3)=-3(x-4) \\ \text{or,}~~ 4y-12=-3x+12 \\ \therefore~~ 3x+4y=24~~\text{(ans.)}$

Solution.

$~x^2+y^2-4x-6y-3=0 \longrightarrow(1)$

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

$~2x+2y~\frac{dy}{dx}-4-6~\frac{dy}{dx}=0 \\ \text{or,}~~ 2\left(x+y~\frac{dy}{dx}-2-3~\frac{dy}{dx}\right)=0\\ \text{or,}~~ (y-3)~\frac{dy}{dx}+(x-2)=0 \\ \text{or,}~~ \frac{dy}{dx}=\frac{2-x}{y-3} \\ \therefore~~ \left[\frac{dy}{dx}\right]_{(2,-1)}=\frac{2-2}{-1-3}=0.$

$\therefore~~$ The equation of tangent $~(2,-1)~$ to the circle (1) is given by 

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

Mtg 45 + 21 Years Jee Main And Iit Jee Advanced Previous Years Solved Papers With Chapterwise Topicwise Solutions Physics, Chemistry, Mathematics ( Set ... Jee Advanced Pyq Question Bank For 2023 Exam )

7. the circle $~x^2+y^2-4x-6y-3=0~$ at $~(2,-1)$

Solution.

$~xy=16 \longrightarrow(1)$

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

$~ x \cdot \frac{dy}{dx}+y \cdot 1=0 \\ \text{or,}~~ \frac{dy}{dx}=-\frac yx. \\ \therefore~~\left[\frac{dy}{dx}\right]_{(-4,-4)}=-\frac{-4}{-4}=-1.$

So, the equation of tangent at $~(-4,-4)~$ to the rectangular hyperbola (1) is given by 

$~~y-(-4)=\left[\frac{dy}{dx}\right]_{(-4,-4)} \left(x-(-4)\right) \\ \text{or,}~~ y+4=-1(x+4) \\ \text{or,}~~ y+4=-x-4 \\ \text{or,}~~ x+y+8=0~~\text{(ans.)}$

8. the rectangular hyperbola $~xy=16~$ at $~(-4,-4).$

Solution.

$~xy=16 \longrightarrow(1)$

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

$~ x \cdot \frac{dy}{dx}+y \cdot 1=0 \\ \text{or,}~~ \frac{dy}{dx}=-\frac yx. \\ \therefore~~\left[\frac{dy}{dx}\right]_{(-4,-4)}=-\frac{-4}{-4}=-1.$

So, the equation of tangent at $~(-4,-4)~$ to the rectangular hyperbola (1) is given by 

$~~y-(-4)=\left[\frac{dy}{dx}\right]_{(-4,-4)} \left(x-(-4)\right) \\ \text{or,}~~ y+4=-1(x+4) \\ \text{or,}~~ y+4=-x-4 \\ \text{or,}~~ x+y+8=0~~\text{(ans.)}$

9. the hyperbola $~~\frac{x^2}{a^2}-\frac{y^2}{b^2}=1~$ at $~(x,y).$

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 both sides of (1) w.r.t. $~x~$, we get

$~b^2 \cdot 2x+a^2 \cdot 2y~p=0 \\ \text{or,}~~ 2(b^2x-a^2y~p)=0 \\ \text{or,}~~ b^2x-a^2y~p=0 \\ \therefore~~ p=\frac{b^2x}{a^2y}$

So, the equation of tangent at $~(x,y)~$ to the hyperbola (1) is 

$~Y-y=\left[\frac{dy}{dx}\right]_{(x,y)}(X-x) \\ \text{or,}~~ Y-y=\frac{b^2x}{a^2y}(X-x)  \\ \text{or,}~~ a^2y \cdot Y-a^2y^2=b^2xX-b^2x^2 \\ \text{or,}~~ b^2x^2-a^2y^2=b^2xX-a^2yY \\ \text{or,}~~a^2b^2=b^2xX-a^2yY ~~[\text{By (1)}]\\ \therefore~~ b^2xX-a^2yY=a^2b^2~~\text{(ans)}$

10. the parabola $~y^2=-36x~$ at point whose ordinate is three times the abscissa.

Solution.

$~y^2=-36x \longrightarrow(1)$

By question, let the coordinates of the required point be $~(h,3h).$

So, the point $~(h,3h)~$ satisfies the equation of the parabola (1).

$\therefore~~ (3h)^2=-36 h \\ \text{or,}~~ 9h^2=-36h \\ \therefore~~ h=\frac{-36}{9}=-4 ~~( h \neq 0).$

$\therefore~~$ The  coordinates of the required point on the parabola (1) is $~(h,3h)=(-4,-12).$

Now, differentiating both sides of (1) w.r.t. $~x,~$ we get

$~\frac{d}{dx}(y^2)=-36 ~\frac{d}{dx}(x) \\ \text{or,}~~ 2y~\frac{dy}{dx}==-36 \\ \text{or,}~~ \frac{dy}{dx}= \frac{-36}{2y}=\frac{-18}{y} \\ \therefore~ \left[\frac{dy}{dx}\right]_{(-4,-12)}=\frac{-18}{-12}=\frac 32=m~(\text{say})$ 

So, the equation of tangent to the parabola (1) at $~(-4,-12)~$ is 

$~ y-(-12)=m[x-(-4)] \\ \text{or,}~~ y+12=\frac 32(x+4) \\ \text{or,}~~ 2y+24=3x+12 \\ \text{or,}~~ 3x-2y=12~~ \text{(ans.)}$

11.  the ellipse $~~x^2+4y^2=25~$ at point whose ordinate is $~2.$

Solution.

$~x^2+4y^2=25 \longrightarrow(1)$

Let the coordinates of the required point on the ellipse is $~(h,2).$

So, the point $~(h,2)~$ satisfies the equation (1).

$\therefore~~ h^2+4 \times 2^2=25 \Rightarrow ~ h=\pm \sqrt{25-16}=\pm{9}=\pm 3.$

So, the coordinates of the required points are $~( \pm 3,2).$

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

$~2x+4 \times 2y~\frac{dy}{dx}=0 \\ \text{or,}~~ 2\left(x+4y~\frac{dy}{dx}\right)=0 \Rightarrow ~ \frac{dy}{dx}=\frac{-x}{4y} \\ \text{So,}~~\left[\frac{dy}{dx}\right]_{(3,2)}=\frac{-3}{4 \times 2}=-\frac 38=m~~\text{(say)}$

$\therefore~~$ The equation of tangent on the ellipse (1) at $~(3,2)~$ is 

$~y-2=m(x-3) \\ \text{or,}~~ y-2=-\frac 38(x-3) \\ \text{or,}~~ 8y-16=-3x+9 \\ \text{or,}~~ 3x+8y=25\rightarrow(2)$

Again, $~\left[\frac{dy}{dx}\right]_{(-3,2)}=-\frac{-3}{4 \times 2}=\frac 38=m'~\text{(say)}$

So, the equation of the tangent on the ellipse (1) at $~(-3,2)~$ is 

$~y-2=m'[x-(-3)] \\ \text{or,}~~ y-2=\frac 38(x+3) \\ \text{or,}~~ 8y-16=3x+9 \\ \therefore~~ 3x-8y+25=0 \rightarrow(3)$

From (2) and (3), we get the required equations of tangents.

12. the circle $~x^2+y^2-6x-2y+6=0~$ at points equidistant from coordinate axes.

Solution.

$x^2+y^2-6x-2y+6=0 \longrightarrow(1)$

Let the required point be $~(a,a)~$ which clearly satisfies equation (1) as it lies on the circle (1).

$~\therefore~~ a^2+a^2-6a-2a+6=0 \\ \text{or,}~~ 2a^2-8a+6=0 \\ \text{or,}~~ 2(a^2-4a+3)=0 \\ \text{or,}~~ a^2-3a-a+3=0 \\ \text{or,}~~ a(a-3)-1(a-3)=0 \\ \text{or,}~~ (a-3)(a-1)=0 \\ \therefore~~ a=3,1.$

So, the coordinates of the required points are $~(3,3)~$ and $~(1,1).$

Now, differentiating (1) w.r.t. to $~x,~$ we get

$~2x+2y~\frac{dy}{dx}-6-2~\frac{dy}{dx}=0 \\ \text{or,}~~ 2\left(x+y\frac{dy}{dx}-3-\frac{dy}{dx}\right)=0 \\ \text{or,}~~ x+y~\frac{dy}{dx}-3-\frac{dy}{dx}=0 \\ \text{or,}~~ \frac{dy}{dx}(y-1)=3-x \\ \therefore~~ \frac{dy}{dx}=\frac{3-x}{y-1}.$

$\therefore~~\left[\frac{dy}{dx}\right]_{(3,3)}=\frac{3-3}{3-1}=0.$

So, at $~(3,3)~$, the tangent is parallel to x-axis and the equation of tangent is $~~y-3=0\rightarrow(2).$

Again, at $~(1,1),~$

$~\left[\frac{dy}{dx}\right]_{(1,1)}=\frac{1-1}{3-1}=0.$

So, the equation of tangent to the circle (1) at $~(1,1)~$ is given by 

$~x-1=\left[\frac{dy}{dx}\right]_{(1,1)}~(y-1) \\ \text{or,}~~ x-1=0 \cdot (y-1) \\ \therefore~~ x-1=0 \rightarrow(3)$

From (2) and (3), we get the required equations of tangents.

13. the curve $~x^{\frac 23}+y^{\frac 23}=a^{\frac 23}~$ at $~(x_1,y_1).$

Solution.

$~x^{2/3}+y^{2/3}=a^{2/3} \longrightarrow(1)$

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

$~\frac 23 x^{\frac 23-1}+\frac 23 y^{\frac 23-1}~\frac{dy}{dx}=0 \\ \text{or,}~~ \frac 23 \left[x^{-1/3}+y^{-1/3}~\frac{dy}{dx}\right]=0 \\ \text{or,}~~ \frac{dy}{dx}=-\frac{x^{-1/3}}{y^{-1/3}}=-\left(\frac yx\right)^{\frac 13}$

Since the point $~(x_1,y_1)~$ lies on the curve (1), so $~ x_1^{2/3}+y_1^{2/3}=a^{2/3} \longrightarrow(2)$

The equation of tangent on the curve (1) at $~(x_1,y_1)~$ is given by 

$~y-y_1=\left[\frac{dy}{dx}\right]_{(x_1,y_1)}(x-x_1) \\ \text{or,}~~ y-y_1=-\left(\frac{y_1}{x_1}\right)^{\frac 13}(x-x_1) \\ \text{or,}~~ x_1^{\frac 13}(y-y_1)=-y_1^{\frac 13}(x-x_1) \\ \text{or,}~~ x_1^{\frac 13}y-x_1^{\frac 13}y_1=-xy_1^{\frac 13}+x_1y_1^{\frac 13} \\ \text{or,}~~ yx_1^{\frac 13}+xy_1^{\frac 13}=x_1y_1^{\frac 13}+x_1^{\frac 13}y_1 \\ \text{or,}~~ yx_1^{\frac 13}+xy_1^{\frac 13}=x_1^{\frac 13}y_1^{\frac 13}(x_1^{\frac 23}+y_1^{\frac 23}) \\ \text{or,}~~ yx_1^{\frac 13}+xy_1^{\frac 13}= x_1^{\frac 13}y_1^{\frac 13} \cdot a^{\frac 23}\\ \text{or,}~~ \frac{yx_1^{\frac 13}}{x_1^{\frac 13}y_1^{\frac 13}}+\frac{xy_1^{\frac 13}}{x_1^{\frac 13}y_1^{\frac 13}}=a^{2/3} \\ \text{or,}~~ yy_1^{-\frac 13}+xx_1^{-\frac 13} =a^{2/3}\\ \text{or,}~~ xx_1^{-\frac 13}+yy_1^{-\frac 13}=a^{\frac 23}~~\text{(ans)}$

14. the ellipse $~x=a\cos\theta,~y=b\sin\theta~$ at $~\theta=\frac{\pi}{3}.$

Solution.

$~x=a\cos\theta \Rightarrow \frac{dx}{d\theta}=-a\sin\theta, \\~~y=b\sin\theta \Rightarrow \frac{dy}{d\theta}=b\cos\theta \\ \therefore~~ \frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}=\frac{b\cos\theta}{-a\sin\theta}=-\frac ba \cot\theta$

$~\left[\frac{dy}{dx}\right]_{\theta =\frac{\pi}{3}}=-\frac ba \cdot \cot \frac{\pi}{3}=-\frac{b}{a\sqrt{3}}.$

At $~\theta=\frac{\pi}{3},~~ x=a\cos(\pi/3)=a \cdot \frac 12=\frac a2.$

At $~\theta=\frac{\pi}{3},~~ y=b\sin(\pi/3)=b \cdot \frac{\sqrt{3}}{2}=\frac{\sqrt{3}b}{2}.$

The equation of tangent at $~\theta=\pi/3,~$ is given by 

$~y-\frac{\sqrt{3}b}{2}=\left[\frac{dy}{dx}\right]_{\theta=\pi/3} \left(x-\frac a2\right) \\ \text{or,}~~ y-\frac{\sqrt{3}b}{2}=-\frac{b}{a\sqrt{3}}(x-a/2) \\ \text{or,}~~ \frac 12(2y-\sqrt{3}b)=\frac{-b}{a\sqrt{3}} \times \frac 12(2x-a) \\ \text{or,}~~ (2y-\sqrt{3}b) \cdot a\sqrt{3}=-b(2x-a) \\ \text{or,}~~ 2\sqrt{3} \cdot ay-3ab=-2bx+ab \\ \text{or,}~~ 2bx+2\sqrt{3}\cdot ay=4ab \\ \text{or,}~~ 2(bx+\sqrt{3} ay)=2 \times 2ab \\ \text{or,}~~ bx+\sqrt{3} ay=2ab \\ \text{or,}~~ \frac{bx}{ab}+\frac{\sqrt{3}ay}{ab}=2 \\ \therefore~~ \frac xa+\frac{\sqrt{3}y}{b}=2~~\text{(ans)}$

15. the curve $~x^3+xy^2-3x^2+4x+5y+2=0~$ at $~(1,-1).$

Solution.

$~x^3+xy^2-3x^2+4x+5y+2=0 \longrightarrow(1)$

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

$~3x^2+1 \cdot y^2+x \cdot 2y~\frac{dy}{dx}-6x+4+5~\frac{dy}{dx}=0 \\ \text{or,}~~ \frac{dy}{dx}(2xy+5)=6x-3x^2-y^2-4 \\ \text{or,}~~ \frac{dy}{dx}=\frac{-3x^2-y^2+6x-4}{2xy+5} \\ \therefore~~ \left[\frac{dy}{dx}\right]_{(1,-1)}=\frac{-3 \cdot 1^2-(-1)^2+6 \cdot 1-4}{2 \cdot 1 \cdot (-1)+5}=-\frac 23.$

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

$~y-(-1)=-\frac 23(x-1) \\ \text{or,}~~ y+1=-\frac 23(x-1) \\ \text{or,}~~ 3y+3=-2x+2 \\ \therefore~~ 2x+3y+1=0~~ \text{(ans)}$