Find the equations of the tangent
Question:

Find the equations of the tangent and the normal to the following curves at the indicated points.

(i) $y=x^{4}-b x^{3}+13 x^{2}-10 x+5$ at $(0,5) \quad$ [NCERT]

(ii) $y=x^{4}-6 x^{3}+13 x^{2}-10 x+5$ at $x=1 \quad$ [NCERT, CBSE 2011]

$\begin{array}{ll}\text { (iii) } y=x^{2} \text { at }(0,0) & \text { [NCERT] }\end{array}$

(iv) $y=2 x^{2}-3 x-1$ at $(1,-2)$

(v) $y^{2}=\frac{x^{3}}{4-x}$ at $(2,-2)$

(vi) $y=x^{2}+4 x+1$ at $x=3 \quad$ [CBSE 2004]

(vii) $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ at $(a \cos \theta, b \sin \theta)$

(viii) $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ at $(a \sec \theta, b \tan \theta)$

(ix) $y^{2}=4 a x$ at $\left(\frac{a}{m^{2}}, \frac{2 a}{m}\right)$

(x) $c^{2}\left(x^{2}+y^{2}\right)=x^{2} y^{2}$ at $\left(\frac{c}{\cos \theta}, \frac{c}{\sin \theta}\right)$

(ix) $x y=c^{2}$ at $\left(c t, \frac{c}{t}\right)$

(xii) $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ at $\left(x_{1}, y_{1}\right)$

(xiii) $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ at $\left(x_{0}, y_{0}\right) \quad$ [NCERT]

(xiv) $x^{\frac{2}{3}}+y^{\frac{2}{3}}=2$ at $(1,1) \quad$ [NCERT]

$(\mathrm{XV}) x^{2}=4 y$ at $(2,1)$

(xvi) $y^{2}=4 x$ at $(1,2) \quad$ [NCERT]

(xvii) $4 x^{2}+9 y^{2}=36$ at $(3 \cos \theta, 2 \sin \theta) \quad$ [CBSE 2011]

(xviii) $y^{2}=4 a x$ at $\left(x_{1}, y_{1}\right) \quad$ [CBSE 2012]

 

(xix) $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ at $(\sqrt{2} a, b) \quad$ [CBSE 2014]

Solution:

(i)

$y=x^{4}-b x^{3}+13 x^{2}-10 x+5$

Differentiating both sides w.r.t. $x$,

$\frac{d y}{d x}=4 x^{3}-3 b x^{2}+26 x-10$

Slope of tangent, $m=\left(\frac{d y}{d x}\right)_{(0,5)}=-10$

Given $\left(x_{1}, y_{1}\right)=(0,5)$

Equation of tangent is,

$y-y_{1}=m\left(x-x_{1}\right)$

$\Rightarrow y-5=-10(x-0)$

$\Rightarrow y-5=-10 x$

$\Rightarrow y+10 x-5=0$

Equation of normal is,

$y-y_{1}=\frac{-1}{m}\left(x-x_{1}\right)$

$\Rightarrow y-5=\frac{1}{10}(x-0)$

$\Rightarrow 10 y-50=x$

$\Rightarrow x-10 y+50=0$

(ii)

$y=x^{4}-6 x^{3}+13 x^{2}-10 x+5$

When $x=1, y=1-6+13-10+5=3$

So, $\left(x_{1}, y_{1}\right)=(1,3)$

Now, $y=x^{4}-6 x^{3}+13 x^{2}-10 x+5$

 

Differentiating both sides w.r.t. $x$,

$\frac{d y}{d x}=4 x^{3}-18 x^{2}+26 x-10$

Slope of tangent, $m=\left(\frac{d y}{d x}\right)_{(1,3)}=4-18+26-10=2$

Equation of tangent is,

$y-y_{1}=2\left(x-x_{1}\right)$

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

$\Rightarrow y-3=2 x-2$

 

$\Rightarrow 2 x-y+1=0$

Equation of normal is,

$y-y_{1}=\frac{-1}{m}\left(x-x_{1}\right)$

$\Rightarrow y-3=\frac{-1}{2}(x-1)$

$\Rightarrow 3 y-6=-x+1$

 

$\Rightarrow x+3 y-7=0$

(iii)

$y=x^{2}$

Differentiating both sides w.r.t. $x$,

$\frac{d y}{d x}=2 x$

Given $\left(x_{1}, y_{1}\right)=(0,0)$

Slope of tangent, $m=\left(\frac{d y}{d x}\right)_{(0,0)}=2(0)=0$

Equation of tangent is,

$y-y_{1}=m\left(x-x_{1}\right)$

$\Rightarrow y-0=0(x-0)$

 

$\Rightarrow y=0$

(iv)

$y=2 x^{2}-3 x-1$

Differentiating both sides w.r.t. $x$,

$\frac{d y}{d x}=4 x-3$

Given $\left(x_{1}, y_{1}\right)=(1,-2)$

Slope of tangent, $m=\left(\frac{d y}{d x}\right)_{(1,-2)}=4-3=1$

Equation of tangent is,

$y-y_{1}=m\left(x-x_{1}\right)$

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

$\Rightarrow y+2=x-1$

 

$\Rightarrow x-y-3=0$

Equation of normal is,

$y-y_{1}=\frac{-1}{m}\left(x-x_{1}\right)$

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

$\Rightarrow y+2=-x+1$

 

$\Rightarrow x+y+1=0$

(v)

$y^{2}=\frac{x^{3}}{4-x}$

Differentiating both sides w.r.t. $x$,

$2 y \frac{d y}{d x}=\frac{(4-x)\left(3 x^{2}\right)-x^{3}(-1)}{(4-x)^{2}}=\frac{12 x^{2}-3 x^{3}+x^{3}}{(4-x)^{2}}=\frac{12 x^{2}-2 x^{3}}{(4-x)^{2}}$

$\frac{d y}{d x}=\frac{12 x^{2}-2 x^{3}}{2 y(4-x)^{2}}$

Given $\left(x_{1}, y_{1}\right)=(2,-2)$

Slope of tangent, $m=\left(\frac{d y}{d x}\right)_{(2,-2)}=\frac{48-16}{-16}=-2$

Equation of tangent is,

$y-y_{1}=m\left(x-x_{1}\right)$

$\Rightarrow y+2=-2(x-2)$

$\Rightarrow y+2=-2 x+4$

 

$\Rightarrow 2 x+y-2=0$

Equation of normal is,

$y-y_{1}=\frac{-1}{m}\left(x-x_{1}\right)$

$\Rightarrow y+2=\frac{1}{2}(x-2)$

$\Rightarrow 2 y+4=x-2$

 

$\Rightarrow x-2 y-6=0$

(vi)

$y=x^{2}+4 x+1$

Differentiating both sides w.r.t. $x$,

$\frac{d y}{d x}=2 x+4$

When $x=3, y=9+12+1=22$

So, $\left(x_{1}, y_{1}\right)=(3,22)$

Slope of tangent, $m=\left(\frac{d y}{d x}\right)_{x=3}=10$

 

Equation of tangent is,

$y-y_{1}=m\left(x-x_{1}\right)$

$\Rightarrow y-22=10(x-3)$

$\Rightarrow y-22=10 x-30$

 

$\Rightarrow 10 x-y-8=0$

Equation of normal is,

$y-y_{1}=\frac{-1}{m}\left(x-x_{1}\right)$

$\Rightarrow y-22=\frac{-1}{10}(x-3)$

$\Rightarrow 10 y-220=-x+3$

 

$\Rightarrow x+10 y-223=0$

(vii)

$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$

Differentiating both sides w.r.t. $x$,

$\Rightarrow \frac{2 x}{a^{2}}+\frac{2 y}{b^{2}} \frac{d y}{d x}=0$

$\Rightarrow \frac{2 y}{b^{2}} \frac{d y}{d x}=\frac{-2 x}{a^{2}}$

 

$\Rightarrow \frac{d y}{d x}=\frac{-x b^{2}}{y a^{2}}$

Slope of tangent, $m=\left(\frac{d y}{d x}\right)_{(a \cos \theta, b \sin \theta)}=\frac{-a \cos \theta\left(b^{2}\right)}{b \sin \theta\left(a^{2}\right)}=\frac{-b \cos \theta}{a \sin \theta}$

Given $\left(x_{1}, y_{1}\right)=(a \cos \theta, b \sin \theta)$

Equation of tangent is,

$y-y_{1}=m\left(x-x_{1}\right)$

$\Rightarrow y-b \sin \theta=\frac{-b \cos \theta}{a \sin \theta}(x-a \cos \theta)$

$\Rightarrow a y \sin \theta-a b \sin ^{2} \theta=-b x \cos \theta+a b \cos ^{2} \theta$

$\Rightarrow b x \cos \theta+a y \sin \theta=a b$

Dividing by ab,

 

$\Rightarrow \frac{x}{a} \cos \theta+\frac{y}{b} \sin \theta=1$

Equation of normal is,

$y-y_{1}=\frac{-1}{m}\left(x-x_{1}\right)$

 

$\Rightarrow y-b \sin \theta=\frac{a \sin \theta}{b \cos \theta}(x-a \cos \theta)$

$\Rightarrow b y \cos \theta-b^{2} \sin \theta \cos \theta=a x \sin \theta-a^{2} \sin \theta \cos \theta$

 

$\Rightarrow a x \sin \theta-b y \cos \theta=\left(a^{2}-b^{2}\right) \sin \theta \cos \theta$

Dividing by $\sin \theta \cos \theta$,

$a x \sec \theta-b y \operatorname{cosec} \theta=\left(a^{2}-b^{2}\right)$

(viii)

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

Differentiating both sides w.r.t. $x$,

$\Rightarrow \frac{2 x}{a^{2}}-\frac{2 y}{b^{2}} \frac{d y}{d x}=0$

$\Rightarrow \frac{2 y}{b^{2}} \frac{d y}{d x}=\frac{2 x}{a^{2}}$

 

$\Rightarrow \frac{d y}{d x}=\frac{x b^{2}}{y a^{2}}$

Slope of tangent, $m=\left(\frac{d y}{d x}\right)_{(a \sec \theta, b \tan \theta)}=\frac{a \sec \theta\left(b^{2}\right)}{b \tan \theta\left(a^{2}\right)}=\frac{b}{a \sin \theta}$

Given $\left(x_{1}, y_{1}\right)=(a \sec \theta, b \tan \theta)$

Equation of tangent is,

$y-y_{1}=m\left(x-x_{1}\right)$

$\Rightarrow y-b \tan \theta=\frac{b}{a \sin \theta}(x-a \sec \theta)$

$\Rightarrow a y \sin \theta-a b \frac{\sin ^{2} \theta}{\cos \theta}=b x-\frac{a b}{\cos \theta}$

$\Rightarrow \frac{a y \sin \theta \cos \theta-a b \sin ^{2} \theta}{\cos \theta}=\frac{b x \cos \theta-a b}{\cos \theta}$

$\Rightarrow a y \sin \theta \cos \theta-a b \sin ^{2} \theta=b x \cos \theta-a b$

$\Rightarrow b x \cos \theta-a y \sin \theta \cos \theta=a b\left(1-\sin ^{2} \theta\right)$

$\Rightarrow b x \cos \theta-a y \sin \theta \cos \theta=a b \cos ^{2} \theta$

Dividing by $a b \cos ^{2} \theta$,

 

$\Rightarrow \frac{x}{a} \sec \theta-\frac{y}{b} \tan \theta=1$

Equation of normal is,

$y-y_{1}=\frac{-1}{m}\left(x-x_{1}\right)$

$\Rightarrow y-b \tan \theta=\frac{-a \sin \theta}{b}(x-a \sec \theta)$

$\Rightarrow y b-b^{2} \tan \theta=-a x \sin \theta+a^{2} \tan \theta$

$\Rightarrow a x \sin \theta+b y=\left(a^{2}+b^{2}\right) \tan \theta$

Dividing by $\tan \theta$,

 

$a x \cos \theta+b y \cot \theta=\left(a^{2}+b^{2}\right)$

(ix)

$y^{2}=4 a x$

Differentiating both sides w.r.t. $x$,

$2 y \frac{d y}{d x}=4 a$

 

$\Rightarrow \frac{d y}{d x}=\frac{2 a}{y}$

Given $\left(x_{1}, y_{1}\right)=\left(\frac{a}{m^{2}}, \frac{2 a}{m}\right)$

Slope of tangent $=\left(\frac{d y}{d x}\right)\left(\frac{a}{m^{2}}, \frac{2 a}{m}\right)=\frac{2 a}{\left(\frac{2 a}{m}\right)}=m$

Equation of tangent is,

$y-y_{1}=m\left(x-x_{1}\right)$

$\Rightarrow y-\frac{2 a}{m}=m\left(x-\frac{a}{m^{2}}\right)$

$\Rightarrow \frac{m y-2 a}{m}=m\left(\frac{m^{2} x-a}{m^{2}}\right)$

$\Rightarrow m y-2 a=m^{2} x-a$

 

$\Rightarrow m^{2} x-m y+a=0$

Equation of normal is,

$y-y_{1}=\frac{1}{\text { Slope of tangent }}\left(x-x_{1}\right)$

$\Rightarrow y-\frac{2 a}{m}=\frac{-1}{m}\left(x-\frac{a}{m^{2}}\right)$

$\Rightarrow \frac{m y-2 a}{m}=\frac{-1}{m}\left(\frac{m^{2} x-a}{m^{2}}\right)$

$\Rightarrow m^{3} y-2 a m^{2}=-m^{2} x+a$

 

$\Rightarrow m^{2} x+m^{3} y-2 a m^{2}-a=0$

(x)

$c^{2}\left(x^{2}+y^{2}\right)=x^{2} y^{2}$

Differentiating both sides w.r.t. $x$,

$\Rightarrow 2 x c^{2}+2 y c^{2} \frac{d y}{d x}=x^{2} 2 y \frac{d y}{d x}+2 x y^{2}$

$\Rightarrow \frac{d y}{d x}\left(2 y c^{2}-2 x^{2} y\right)=2 x y^{2}-2 x c^{2}$

$\Rightarrow \frac{d y}{d x}=\frac{x y^{2}-x c^{2}}{y c^{2}-x^{2} y}$

Slope of tangent, $m=\left(\frac{d y}{d x}\right)\left(\frac{c}{\cos \theta}, \frac{c}{\sin \theta}\right)$

$=\frac{\frac{c^{3}}{\cos \theta \sin ^{2} \theta}-\frac{c^{3}}{\cos \theta}}{\frac{c^{3}}{\sin \theta}-\frac{c^{3}}{\cos ^{2} \theta \sin \theta}}$

$=\frac{\frac{1-\sin ^{2} \theta}{\cos \theta \sin ^{2} \theta}}{\frac{\cos ^{2} \theta-1}{\cos ^{2} \theta \sin \theta}}$

$=\frac{\cos ^{2} \theta}{\cos \theta \sin ^{2} \theta} \times \frac{\cos ^{2} \theta \sin \theta}{-\sin ^{2} \theta}$

$=\frac{-\cos ^{3} \theta}{\sin ^{3} \theta}$

Given $\left(x_{1}, y_{1}\right)=\left(\frac{c}{\cos \theta}, \frac{c}{\sin \theta}\right)$

Equation of tangent is,

$y-y_{1}=m\left(x-x_{1}\right)$

$\Rightarrow y-\frac{c}{\sin \theta}=\frac{-\cos ^{3} \theta}{\sin ^{3} \theta}\left(x-\frac{c}{\cos \theta}\right)$

$\Rightarrow \frac{y \sin \theta-c}{\sin \theta}=\frac{-\cos ^{3} \theta}{\sin ^{3} \theta}\left(\frac{x \cos \theta-c}{\cos \theta}\right)$

$\Rightarrow \sin ^{2} \theta(y \sin \theta-c)=-\cos ^{2} \theta(x \cos \theta-c)$

$\Rightarrow y \sin ^{3} \theta-c \sin ^{2} \theta=-x \cos ^{3} \theta+c \cos ^{2} \theta$

$\Rightarrow x \cos ^{3} \theta+y \sin ^{3} \theta=c\left(\sin ^{2} \theta+\cos ^{2} \theta\right)$

$\Rightarrow x \cos ^{3} \theta+y \sin ^{3} \theta=c$

Equation of normal is,

$y-y_{1}=\frac{-1}{m}\left(x-x_{1}\right)$

$\Rightarrow y-\frac{c}{\sin \theta}=\frac{\sin ^{3} \theta}{\cos ^{3} \theta}\left(x-\frac{c}{\cos \theta}\right)$

$\Rightarrow \cos ^{3} \theta\left(y-\frac{c}{\sin \theta}\right)=\sin ^{3} \theta\left(x-\frac{c}{\cos \theta}\right)$

$\Rightarrow y \cos ^{3} \theta-\frac{c \cos ^{3} \theta}{\sin \theta}=x \sin ^{3} \theta-\frac{c \sin ^{3} \theta}{\cos \theta}$

$\Rightarrow x \sin ^{3} \theta-y \cos ^{3} \theta=\frac{c \sin ^{3} \theta}{\cos \theta}-\frac{c \cos ^{3} \theta}{\sin \theta}$

$\Rightarrow x \sin ^{3} \theta-y \cos ^{3} \theta=c\left(\frac{\sin ^{4} \theta-\cos ^{4} \theta}{\cos \theta \sin \theta}\right)$

$\Rightarrow x \sin ^{3} \theta-y \cos ^{3} \theta=c\left(\frac{\sin ^{4} \theta-\cos ^{4} \theta}{\cos \theta \sin \theta}\right)$

$\Rightarrow x \sin ^{3} \theta-y \cos ^{3} \theta=c\left[\frac{\left(\sin ^{2} \theta+\cos ^{2} \theta\right)\left(\sin ^{2} \theta-\cos ^{2} \theta\right)}{\cos \theta \sin \theta}\right]$

$\Rightarrow \sin ^{3} \theta-y \cos ^{3} \theta=2 c\left[\frac{-\left(\cos ^{2} \theta-\sin ^{2} \theta\right)}{2 \cos \theta \sin \theta}\right.$

]$\Rightarrow \sin ^{3} \theta-y \cos ^{3} \theta=2 c\left[\frac{-\cos (2 \theta)}{\sin (2 \theta)}\right]$

$\Rightarrow \sin ^{3} \theta-y \cos ^{3} \theta=-2 c \cot (2 \theta)$

$\Rightarrow \sin ^{3} \theta-y \cos ^{3} \theta+2 c \cot (2 \theta)=0$

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