Question

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

(i) y = x⁴ − bx³ + 13x² − 10x + 5 at (0, 5) [NCERT]
(ii) y = x⁴ − 6x³ + 13x² − 10x + 5 at x = 1 [NCERT, CBSE 2011]
(iii) y = x² at (0, 0) [NCERT]
(iv) y = 2x² − 3x − 1 at (1, −2)
(v) $y^2=\frac{x^3}{4-x}\mathrm{at}\left(2,-2\right)$
(vi) y = x² + 4x + 1 at x = 3 [CBSE 2004]
(vii) $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\mathrm{at}\left(a\cos \theta ,b\sin \theta \right)$
(viii) $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\mathrm{at}\left(a\sec \theta ,b\tan \theta \right)$
(ix) y² = 4ax at $\left(\frac{a}{m^2},\frac{2a}{m}\right)$
(x) $c^2\left(x^2+y^2\right)=x^2{y}^2\mathrm{at}\left(\frac{c}{\cos \theta },\frac{c}{\sin \theta }\right)$
(ix) xy = c² at $\left(ct,\frac{c}{t}\right)$
(xii) $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\mathrm{at}\left(x_1,y_1\right)$
(xiii) $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\mathrm{at}\left(x_0,y_0\right)$ [NCERT]
(xiv) $x^{\frac{2}{3}}+y^{\frac{2}{3}}$ = 2 at (1, 1) [NCERT]
(xv) x² = 4y at (2, 1)
(xvi) y² = 4x at (1, 2) [NCERT]
(xvii) 4x² + 9y² = 36 at (3cosθ, 2sinθ) [CBSE 2011]
(xviii) y² = 4ax at (x₁, y₁) [CBSE 2012]
(xix) $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\mathrm{at}\left(\sqrt{2}a,b\right)$ [CBSE 2014]

Answer 1

(i)
$$\begin{gathered} \mathit{\text{y}}\text{=}{\mathit{\text{x}}}^4-b{x}^3+13x^2-10x+5 \\ \text{Differentiating both sides w.r.t.}x, \\ \frac{dy}{dx}=4x^3-3b{x}^2+26x-10 \\ \text{Slope of tangent,}\mathit{\text{m}}\text{=}{\left(\frac{dy}{dx}\right)}_{\left(0,5\right)}\text{=-10} \\ \mathrm{Given}\left(x_1,y_1\right)=\left(0,5\right) \\ \text{Equation of tangent is,} \\ y-y_1=m\left(x-x_1\right) \\ \Rightarrow y-5=-10\left(x-0\right) \\ \Rightarrow y-5=-10x \\ \Rightarrow y+10x-5=0 \end{gathered}$$

$$\begin{gathered} \text{Equation of normal is,} \\ y-y_1=\frac{-1}{m}\left(x-x_1\right) \\ \Rightarrow y-5=\frac{1}{10}\left(x-0\right) \\ \Rightarrow 10y-50=x \\ \Rightarrow x-10y+50=0 \end{gathered}$$

(ii)
$$\begin{gathered} \mathit{\text{y}}\text{=}{\mathit{\text{x}}}^4-6x^3+13x^2-10x+5 \\ \mathrm{When}x=1,y=1-6+13-10+5=3 \\ \mathrm{So},\left(x_1,y_1\right)=\left(1,3\right) \\ \mathrm{Now},\mathit{\text{y}}\text{=}{\mathit{\text{x}}}^4-6x^3+13x^2-10x+5 \\ \text{Differentiating both sides w.r.t.}x, \\ \frac{dy}{dx}=4x^3-18x^2+26x-10 \\ \text{Slope of tangent,}\mathit{\text{m}}\text{=}{\left(\frac{dy}{dx}\right)}_{\left(1,3\right)}\text{=4-18+}26-10=2 \\ \text{Equation of tangent is,} \\ y-y_1=2\left(x-x_1\right) \\ \Rightarrow y-3=2\left(x-1\right) \\ \Rightarrow y-3=2x-2 \\ \Rightarrow 2x-y+1=0 \end{gathered}$$

$$\begin{gathered} \text{Equation of normal is,} \\ y-y_1=\frac{-1}{m}\left(x-x_1\right) \\ \Rightarrow y-3=\frac{-1}{2}\left(x-1\right) \\ \Rightarrow 3y-6=-x+1 \\ \Rightarrow x+3y-7=0 \end{gathered}$$

(iii)
$$\begin{gathered} \mathit{\text{y}}\text{=}{\mathit{\text{x}}}^2 \\ \text{Differentiating both sides w.r.t.}x, \\ \frac{dy}{dx}=2x \\ \mathrm{Given}\left(x_1,y_1\right)=\left(0,0\right) \\ \text{Slope of tangent,}\mathit{\text{m}}\text{=}{\left(\frac{dy}{dx}\right)}_{\left(0,0\right)}\text{=2}\left(0\right)\text{=0} \\ \text{Equation of tangent is,} \\ y-y_1=m\left(x-x_1\right) \\ \Rightarrow y-0=0\left(x-0\right) \\ \Rightarrow y=0 \end{gathered}$$

$$\begin{gathered} \text{Equation of normal is,} \\ \Rightarrow y-y_1=\frac{-1}{m}\left(x-x_1\right) \\ \Rightarrow y-0=\frac{-1}{0}\left(x-0\right) \\ \Rightarrow x=0 \end{gathered}$$

(iv)
$$\begin{gathered} \mathit{\text{y}}\text{=2}{\mathit{\text{x}}}^2\text{-3}\mathit{\text{x}}\text{-1} \\ \text{Differentiating both sides w.r.t.}x, \\ \frac{dy}{dx}=4x-3 \\ \mathrm{Given}\left(x_1,y_1\right)=\left(1,-2\right) \\ \text{Slope of tangent,}\mathit{\text{m}}\text{=}{\left(\frac{dy}{dx}\right)}_{\left(1,-2\right)}\text{=4-3=1} \\ \text{Equation of tangent is,} \\ y-y_1=m\left(x-x_1\right) \\ \Rightarrow y+2=1\left(x-1\right) \\ \Rightarrow y+2=x-1 \\ \Rightarrow x-y-3=0 \end{gathered}$$

$$\begin{gathered} \text{Equation of normal is,} \\ y-y_1=\frac{-1}{m}\left(x-x_1\right) \\ \Rightarrow y+2=-1\left(x-1\right) \\ \Rightarrow y+2=-x+1 \\ \Rightarrow x+y+1=0 \end{gathered}$$

(v)
$$\begin{gathered} {\text{y}}^2\text{=}\frac{x^3}{4-x} \\ \text{Differentiating both sides w.r.t.}x, \\ 2y\frac{dy}{dx}=\frac{\left(4-x\right)\left(3x^2\right)-x^3\left(-1\right)}{{\left(4-x\right)}^2}=\frac{12x^2-3x^3+x^3}{{\left(4-x\right)}^2}=\frac{12x^2-2x^3}{{\left(4-x\right)}^2} \\ \frac{dy}{dx}=\frac{12x^2-2x^3}{2y{\left(4-x\right)}^2} \\ \mathrm{Given}\left(x_1,y_1\right)=\left(2,-2\right) \\ \text{Slope of tangent,}\mathit{\text{m}}\text{=}{\left(\frac{dy}{dx}\right)}_{\left(2,-2\right)}\text{=}\frac{48-16}{-16}\text{=-2} \\ \text{Equation of tangent is,} \\ y-y_1=m\left(x-x_1\right) \\ \Rightarrow y+2=-2\left(x-2\right) \\ \Rightarrow y+2=-2x+4 \\ \Rightarrow 2x+y-2=0 \end{gathered}$$

$$\begin{gathered} \text{Equation of normal is,} \\ y-y_1=\frac{-1}{m}\left(x-x_1\right) \\ \Rightarrow y+2=\frac{1}{2}\left(x-2\right) \\ \Rightarrow 2y+4=x-2 \\ \Rightarrow x-2y-6=0 \end{gathered}$$

(vi)
$$\begin{gathered} y=x^2+4x+1 \\ \text{Differentiating both sides w.r.t.}x, \\ \frac{dy}{dx}=2x+4 \\ \text{When}\mathit{\text{x}}\text{=3,}y=9+12+1=22 \\ \mathrm{So},\left(x_1,y_1\right)=\left(3,22\right) \\ \text{Slope of tangent,}\mathit{\text{m}}\text{=}{\left(\frac{dy}{dx}\right)}_{x=3}\text{=10} \\ \text{Equation of tangent is,} \\ y-y_1=m\left(x-x_1\right) \\ \Rightarrow y-22=10\left(x-3\right) \\ \Rightarrow y-22=10x-30 \\ \Rightarrow 10x-y-8=0 \end{gathered}$$

$$\begin{gathered} \text{Equation of normal is,} \\ y-y_1=\frac{-1}{m}\left(x-x_1\right) \\ \Rightarrow y-22=\frac{-1}{10}\left(x-3\right) \\ \Rightarrow 10y-220=-x+3 \\ \Rightarrow x+10y-223=0 \end{gathered}$$

(vii)
$$\begin{gathered} \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \\ \text{Differentiating both sides w.r.t.}x, \\ \Rightarrow \frac{2x}{a^2}+\frac{2y}{b^2}\frac{dy}{dx}=0 \\ \Rightarrow \frac{2y}{b^2}\frac{dy}{dx}=\frac{-2x}{a^2} \\ \Rightarrow \frac{dy}{dx}=\frac{-x{b}^2}{y{a}^2} \\ \text{Slope of tangent,}\mathit{\text{m}}\text{=}{\left(\frac{dy}{dx}\right)}_{\left(a\cos \theta ,b\sin \theta \right)}\text{=}\frac{-a\cos \theta \left(b^2\right)}{b\sin \theta \left(a^2\right)}\text{=}\frac{-b\cos \theta }{a\sin \theta } \\ \mathrm{Given}\left(x_1,y_1\right)=\left(a\cos \theta ,b\sin \theta \right) \\ \text{Equation of tangent is,} \\ y-y_1=m\left(x-x_1\right) \\ \Rightarrow y-b\sin \theta =\frac{-b\cos \theta }{a\sin \theta }\left(x-a\cos \theta \right) \\ \Rightarrow ay\sin \theta -ab{\sin }^2\theta =-bx\cos \theta +ab{\cos }^2\theta \\ \Rightarrow bx\cos \theta +ay\sin \theta =ab \\ \text{Dividing by ab,} \\ \Rightarrow \frac{x}{a}\cos \theta +\frac{y}{b}\sin \theta =1 \end{gathered}$$

$$\begin{gathered} \text{Equation of normal is,} \\ y-y_1=\frac{-1}{m}\left(x-x_1\right) \\ \Rightarrow y-b\sin \theta =\frac{a\sin \theta }{b\cos \theta }\left(x-a\cos \theta \right) \\ \Rightarrow by\cos \theta -b^2\sin \theta \cos \theta =ax\sin \theta -a^2\sin \theta \cos \theta \\ \Rightarrow ax\sin \theta -by\cos \theta =\left(a^2-b^2\right)\sin \theta \cos \theta \\ \text{Dividing by}\sin \theta \cos \theta , \\ ax\sec \theta -by\mathrm{cosec}\theta =\left(a^2-b^2\right) \end{gathered}$$

(viii)
$$\begin{gathered} \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \\ \text{Differentiating both sides w.r.t.}x, \\ \Rightarrow \frac{2x}{a^2}-\frac{2y}{b^2}\frac{dy}{dx}=0 \\ \Rightarrow \frac{2y}{b^2}\frac{dy}{dx}=\frac{2x}{a^2} \\ \Rightarrow \frac{dy}{dx}=\frac{x{b}^2}{y{a}^2} \\ \text{Slope of tangent,}\mathit{\text{m}}\text{=}{\left(\frac{dy}{dx}\right)}_{\left(a\sec \theta ,b\tan \theta \right)}\text{=}\frac{a\sec \theta \left(b^2\right)}{b\tan \theta \left(a^2\right)}\text{=}\frac{b}{a\sin \theta } \\ \mathrm{Given}\left(x_1,y_1\right)=\left(a\sec \theta ,b\tan \theta \right) \\ \text{Equation of tangent is,} \\ y-y_1=m\left(x-x_1\right) \\ \Rightarrow y-b\tan \theta =\frac{b}{a\sin \theta }\left(x-a\sec \theta \right) \\ \Rightarrow ay\sin \theta -ab\frac{{\sin }^2\theta }{\cos \theta }=bx-\frac{ab}{\cos \theta } \\ \Rightarrow \frac{ay\sin \theta \cos \theta -ab{\sin }^2\theta }{\cos \theta }=\frac{bx\cos \theta -ab}{\cos \theta } \\ \Rightarrow ay\sin \theta \cos \theta -ab{\sin }^2\theta =bx\cos \theta -ab \\ \Rightarrow bx\cos \theta -ay\sin \theta \cos \theta =ab\left(1-{\sin }^2\theta \right) \\ \Rightarrow bx\cos \theta -ay\sin \theta \cos \theta =ab{\cos }^2\theta \\ \text{Dividing by}\mathit{\text{ab}}{\cos }^2\theta \text{,} \\ \Rightarrow \frac{x}{a}\sec \theta -\frac{y}{b}\tan \theta =1 \end{gathered}$$

$$\begin{gathered} \text{Equation of normal is,} \\ y-y_1=\frac{-1}{m}\left(x-x_1\right) \\ \Rightarrow y-b\tan \theta =\frac{-a\sin \theta }{b}\left(x-a\sec \theta \right) \\ \Rightarrow yb-b^2\tan \theta =-ax\sin \theta +a^2\tan \theta \\ \Rightarrow ax\sin \theta +by=\left(a^2+b^2\right)\tan \theta \\ \text{Dividing by tan}\theta , \\ ax\cos \theta +by\cot \theta =\left(a^2+b^2\right) \end{gathered}$$

(ix)
$$\begin{gathered} {\mathit{\text{y}}}^{\mathit{2}}\text{=4}\mathit{\text{ax}} \\ \text{Differentiating both sides w.r.t.}x, \\ 2y\frac{dy}{dx}=4a \\ \Rightarrow \frac{dy}{dx}=\frac{2a}{y} \\ \mathrm{Given}\left(x_1,y_1\right)=\left(\frac{a}{m^2},\frac{2a}{m}\right) \\ \text{Slope of tangent =}{\left(\frac{dy}{dx}\right)}_{\left(\frac{a}{m^2},\frac{2a}{m}\right)}\text{=}\frac{2a}{\left(\frac{2a}{m}\right)}\text{=}\mathit{\text{m}} \\ \text{Equation of tangent is,} \\ y-y_1=m\left(x-x_1\right) \\ \Rightarrow y-\frac{2a}{m}=m\left(x-\frac{a}{m^2}\right) \\ \Rightarrow \frac{my-2a}{m}=m\left(\frac{m^{2}x-a}{m^2}\right) \\ \Rightarrow my-2a=m^{2}x-a \\ \Rightarrow m^{2}x-my+a=0 \end{gathered}$$

$$\begin{gathered} \text{Equation of normal is,} \\ y-y_1=\frac{1}{\mathrm{Slope}\mathrm{of}\mathrm{tangent}}\left(x-x_1\right) \\ \Rightarrow y-\frac{2a}{m}=\frac{-1}{m}\left(x-\frac{a}{m^2}\right) \\ \Rightarrow \frac{my-2a}{m}=\frac{-1}{m}\left(\frac{m^{2}x-a}{m^2}\right) \\ \Rightarrow m^{3}y-2a{m}^2=-m^{2}x+a \\ \Rightarrow m^{2}x+m^{3}y-2a{m}^2-a=0 \end{gathered}$$

(x)
$$\begin{gathered} c^2\left(x^2+y^2\right)=x^2{y}^2 \\ \text{Differentiating both sides w.r.t.}x, \\ \Rightarrow 2x{c}^2+2y{c}^2\frac{dy}{dx}=x^{2}2y\frac{dy}{dx}+2x{y}^2 \\ \Rightarrow \frac{dy}{dx}\left(2y{c}^2-2x^{2}y\right)=2x{y}^2-2x{c}^2 \\ \Rightarrow \frac{dy}{dx}=\frac{x{y}^2-x{c}^2}{y{c}^2-x^{2}y} \\ \text{Slope of tangent,}\mathit{\text{m}}\text{=}{\left(\frac{dy}{dx}\right)}_{\left(\frac{c}{\cos \theta },\frac{c}{\sin \theta }\right)} \\ \text{=}\frac{{\displaystyle \frac{c^3}{\cos \theta {\sin }^2\theta }}-{\displaystyle \frac{c^3}{\cos \theta }}}{{\displaystyle \frac{c^3}{\sin \theta }}-{\displaystyle \frac{c^3}{{\cos }^2\theta \sin \theta }}} \\ =\frac{{\displaystyle \frac{1-{\sin }^2\theta }{\cos \theta {\sin }^2\theta }}}{{\displaystyle \frac{{\cos }^2\theta -1}{{\cos }^2\theta \sin \theta }}} \\ =\frac{co{s}^2\theta }{\cos \theta {\sin }^2\theta }\times \frac{{\cos }^2\theta \sin \theta }{-{\sin }^2\theta } \\ =\frac{-{\cos }^3\theta }{{\sin }^3\theta } \\ \mathrm{Given}\left(x_1,y_1\right)=\left(\frac{c}{\cos \theta },\frac{c}{\sin \theta }\right) \\ \text{Equation of tangent is,} \\ y-y_1=m\left(x-x_1\right) \\ \Rightarrow y-\frac{c}{\sin \theta }=\frac{-{\cos }^3\theta }{{\sin }^3\theta }\left(x-\frac{c}{\cos \theta }\right) \\ \Rightarrow \frac{y\sin \theta -c}{\sin \theta }=\frac{-{\cos }^3\theta }{{\sin }^3\theta }\left(\frac{x\cos \theta -c}{\cos \theta }\right) \\ \Rightarrow {\sin }^2\theta \left(y\sin \theta -c\right)=-{\cos }^2\theta \left(x\cos \theta -c\right) \\ \Rightarrow y{\sin }^3\theta -c{\sin }^2\theta =-x{\cos }^3\theta +c{\cos }^2\theta \\ \Rightarrow x{\cos }^3\theta +y{\sin }^3\theta =c\left(si{n}^2\theta +{\cos }^2\theta \right) \\ \Rightarrow x{\cos }^3\theta +y{\sin }^3\theta =c \end{gathered}$$

$$\begin{gathered} \text{Equation of normal is,} \\ y-y_1=\frac{-1}{m}\left(x-x_1\right) \\ \Rightarrow y-\frac{c}{\sin \mathit{}\theta }=\frac{{\sin }^3\theta }{{\cos }^3\theta }\left(x-\frac{c}{\mathit{cos}\mathit{}\theta }\right) \\ \Rightarrow {\cos }^3\theta \left(y-\frac{c}{\sin \mathit{}\theta }\right)={\sin }^3\theta \left(x-\frac{c}{\cos \mathit{}\theta }\right) \\ \Rightarrow y{\cos }^3\theta -\frac{c{\cos }^3\theta }{\sin \theta }=x{\sin }^3\theta -\frac{c{\sin }^3\theta }{\cos \theta } \\ \Rightarrow x{\sin }^3\theta -y{\cos }^3\theta =\frac{c{\sin }^3\theta }{\cos \theta }-\frac{c{\cos }^3\theta }{\sin \theta } \\ \Rightarrow x{\sin }^3\theta -y{\cos }^3\theta =c\left(\frac{{\sin }^4\theta -{\cos }^4\theta }{\cos \theta \sin \theta }\right) \\ \Rightarrow x{\sin }^3\theta -y{\cos }^3\theta =c\left[\frac{\left({\sin }^2\theta +{\cos }^2\theta \right)\left({\sin }^2\theta -{\cos }^2\theta \right)}{\cos \theta \sin \theta }\right] \\ \Rightarrow {\sin }^3\theta -y{\cos }^3\theta =2c\left[\frac{-\left({\cos }^2\theta -{\sin }^2\theta \right)}{2\cos \theta \sin \theta } \\ \right] \\ \Rightarrow {\sin }^3\theta -y{\cos }^3\theta =2c\left[\frac{-\cos \left(2\theta \right)}{\sin \left(2\theta \right)}\right] \\ \Rightarrow {\sin }^3\theta -y{\cos }^3\theta =-2ccot\left(2\theta \right) \\ \Rightarrow {\sin }^3\theta -y{\cos }^3\theta +2ccot\left(2\theta \right)=0 \end{gathered}$$

(xi)
$$\begin{gathered} {\mathit{\text{xy=c}}}^{\mathit{2}} \\ \text{Differentiating both sides w.r.t.}x, \\ x\frac{dy}{dx}+y=0 \\ \Rightarrow \frac{dy}{dx}=\frac{-y}{x} \\ \mathrm{Given}\left(x_1,y_1\right)=\left(ct,\frac{c}{t}\right) \\ \text{Slope of tangent,}\mathit{\text{m}}\text{=}{\left(\frac{dy}{dx}\right)}_{\left(ct,\frac{c}{t}\right)}\text{=}\frac{-\frac{c}{t}}{ct}\text{=}\frac{-1}{t^2} \\ \text{Equation of tangent is,} \\ y-y_1=m\left(x-x_1\right) \\ \Rightarrow y-\frac{c}{t}=\frac{-1}{t^2}\left(x-ct\right) \\ \Rightarrow \frac{yt-c}{t}=\frac{-1}{t^2}\left(x-ct\right) \\ \Rightarrow y{t}^2-ct=-x+ct \\ \Rightarrow x+y{t}^2=2ct \end{gathered}$$

$$\begin{gathered} \text{Equation of normal is,} \\ y-y_1=\frac{-1}{m}\left(x-x_1\right) \\ \Rightarrow y-\frac{c}{t}={\mathit{\text{t}}}^{\mathit{2}}\left(x-ct\right) \\ \Rightarrow yt-c=t^{3}x-c{t}^4 \\ \Rightarrow x{t}^3-yt=c{t}^4-c \end{gathered}$$

(xii)
$$\begin{gathered} \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \\ \text{Differentiating both sides w.r.t.}x, \\ \frac{2x}{a^2}+\frac{2y}{b^2}\frac{dy}{dx}=0 \\ \Rightarrow \frac{2y}{b^2}\frac{dy}{dx}=\frac{-2x}{a^2} \\ \Rightarrow \frac{dy}{dx}=\frac{-x{b}^2}{y{a}^2} \\ \text{Slope of tangent,}\mathit{\text{m}}\text{=}{\left(\frac{dy}{dx}\right)}_{\left(x_1,y_1\right)}\text{=}\frac{-x_1{b}^2}{y_1{a}^2} \\ \text{Equation of tangent is,} \\ y-y_1=m\left(x-x_1\right) \\ \Rightarrow y-y_1=\frac{-x_1{b}^2}{y_1{a}^2}\left(x-x_1\right) \\ \Rightarrow y{y}_1{a}^2-{y_1}^2{a}^2=-x{x}_1{b}^2+{x_1}^2{b}^2 \\ \Rightarrow x{x}_1{b}^2+y{y}_1{a}^2={x_1}^2{b}^2+{y_1}^2{a}^2...\left(1\right) \\ \text{Since}\left(x_1,y_1\right)\text{lies on the given curve.Therefore,} \\ \frac{{x_1}^2}{a^2}+\frac{{y_1}^2}{b^2}=1 \\ \Rightarrow \frac{{x_1}^2{b}^2+{y_1}^2{a}^2}{a^2{b}^2}=1 \\ \Rightarrow {x_1}^2{b}^2+{y_1}^2{a}^2=a^2{b}^2 \\ \text{Substituting this in (1), we get} \\ x{x}_1{b}^2+y{y}_1{a}^2=a^2{b}^2 \\ {\text{Dividing this by a}}^2{\text{b}}^2\text{,} \\ \frac{x{x}_1}{a^2}+\frac{y{y}_1}{b^2}=1 \end{gathered}$$

$$\begin{gathered} \text{Equation of normal is,} \\ y-y_1=m\left(x-x_1\right) \\ \Rightarrow y-y_1=\frac{y_1{a}^2}{x_1{b}^2}\left(x-x_1\right) \\ \Rightarrow y{x}_1{b}^2-x_1{y}_1{b}^2=x{y}_1{a}^2-x_1{y}_1{a}^2 \\ \Rightarrow x{y}_1{a}^2-y{x}_1{b}^2=x_1{y}_1{a}^2-x_1{y}_1{b}^2 \\ \Rightarrow x{y}_1{a}^2-y{x}_1{b}^2=x_1{y}_1\left(a^2-b^2\right) \\ \text{Dividing by}{x}_1{y}_1 \\ \frac{a^{2}x}{x_1}-\frac{b^{2}y}{y_1}=a^2-b^2 \end{gathered}$$

(xiii)
$$\begin{gathered} \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \\ \text{Differentiating both sides w.r.t.}x, \\ \frac{2x}{a^2}-\frac{2y}{b^2}\frac{dy}{dx}=0 \\ \Rightarrow \frac{2y}{b^2}\frac{dy}{dx}=\frac{2x}{a^2} \\ \Rightarrow \frac{dy}{dx}=\frac{x{b}^2}{y{a}^2} \\ \text{Slope of tangent,}\mathit{\text{m}}\text{=}{\left(\frac{dy}{dx}\right)}_{\left(x_0,y_0\right)}\text{=}\frac{x_0{b}^2}{y_0{a}^2} \\ \text{Equation of tangent is,} \\ y-y_1=m\left(x-x_1\right) \\ \Rightarrow y-y_0=\frac{x_0{b}^2}{y_0{a}^2}\left(x-x_0\right) \\ \Rightarrow y{y}_0{a}^2-{y_0}^2{a}^2=x{x}_0{b}^2-{x_0}^2{b}^2 \\ x{x}_0{b}^2-y{y}_0{a}^2={x_0}^2{b}^2-{y_0}^2{a}^2...\left(1\right) \\ \text{Since}\left(x_0,y_0\right)\text{lies on the given curve,} \\ \Rightarrow \frac{{x_0}^2}{a^2}-\frac{{y_0}^2}{b^2}=1 \\ \Rightarrow {x_0}^2{b}^2-{y_0}^2{a}^2=a^2{b}^2 \\ \text{Substituting this in (1), we get} \\ \Rightarrow x{x}_0{b}^2-y{y}_0{a}^2=a^2{b}^2 \\ \text{Dividing this by}{a}^2{b}^2 \\ \frac{x{x}_0}{a^2}-\frac{y{y}_0}{b^2}=1 \end{gathered}$$

$$\begin{gathered} \text{Equation of normal is,} \\ y-y_1=m\left(x-x_1\right) \\ \Rightarrow y-y_0=\frac{-y_0{a}^2}{x_0{b}^2}\left(x-x_0\right) \\ \Rightarrow y{x}_0{b}^2-x_0{y}_0{b}^2=-x{y}_0{a}^2+x_0{y}_0{a}^2 \\ \Rightarrow x{y}_0{a}^2+y{x}_0{b}^2=x_0{y}_0{a}^2+x_0{y}_0{b}^2 \\ \Rightarrow x{y}_0{a}^2+y{x}_0{b}^2=x_0{y}_0\left(a^2+b^2\right) \\ \text{Dividing by}{x}_0{y}_0 \\ \frac{a^{2}x}{x_0}+\frac{b^{2}y}{y_0}=a^2+b^2 \end{gathered}$$

(xiv)
$$\begin{gathered} x^{\frac{2}{3}}+y^{\frac{2}{3}}=2 \\ \text{Differentiating both sides w.r.t.}x, \\ \frac{2}{3}{x}^{\frac{-1}{3}}+\frac{2}{3}{y}^{\frac{-1}{3}}\frac{dy}{dx}=0 \\ \Rightarrow \frac{dy}{dx}=\frac{-x^{\frac{-1}{3}}}{y^{\frac{-1}{3}}}=\frac{-y^{{\displaystyle \frac{1}{3}}}}{x^{{\displaystyle \frac{1}{3}}}} \\ \text{Slope of tangent,}\mathit{\text{m}}\text{=}{\left(\frac{dy}{dx}\right)}_{\left(1,1\right)}\text{=}\frac{-1}{1}\text{=-1} \\ \mathrm{Given}\left(x_1,y_1\right)=\left(1,1\right) \\ \text{Equation of tangent is,} \\ y-y_1=m\left(x-x_1\right) \\ \Rightarrow y-1=-1\left(x-1\right) \\ \Rightarrow y-1=-x+1 \\ \Rightarrow x+y-2=0 \end{gathered}$$

$$\begin{gathered} \text{Equation of normal is,} \\ y-y_1=\frac{-1}{m}\left(x-x_1\right) \\ \Rightarrow y-1=1\left(x-1\right) \\ \Rightarrow y-1=x-1 \\ \Rightarrow y-x=0 \end{gathered}$$

(xv)
$$\begin{gathered} x^2=4y \\ \text{Differentiating both sides w.r.t.}x, \\ 2x=4\frac{dy}{dx} \\ \Rightarrow \frac{dy}{dx}=\frac{x}{2} \\ \text{Slope of tangent,}\mathit{\text{m}}\text{=}{\left(\frac{dy}{dx}\right)}_{\left(2,1\right)}\text{=}\frac{2}{2}\text{=1} \\ \mathrm{Given}\left(x_1,y_1\right)=\left(2,1\right) \\ \text{Equation of tangent is,} \\ y-y_1=m\left(x-x_1\right) \\ \Rightarrow y-1=1\left(x-2\right) \\ \Rightarrow y-1=x-2 \\ \Rightarrow x-y-1=0 \end{gathered}$$

$$\begin{gathered} \text{Equation of normal is,} \\ y-y_1=\frac{-1}{m}\left(x-x_1\right) \\ \Rightarrow y-1=-1\left(x-2\right) \\ \Rightarrow y-1=-x+2 \\ \Rightarrow x+y-3=0 \end{gathered}$$

(xvi)
$$\begin{gathered} y^2=4x \\ \text{Differentiating both sides w.r.t.}x, \\ 2y\frac{dy}{dx}=4 \\ \Rightarrow \frac{dy}{dx}=\frac{2}{y} \\ \text{Slope of tangent,}\mathit{\text{m}}\text{=}{\left(\frac{dy}{dx}\right)}_{\left(1,2\right)}\text{=}\frac{2}{2}\text{=1} \\ \mathrm{Given}\left(x_1,y_1\right)=\left(1,2\right) \\ \text{Equation of tangent is,} \\ y-y_1=m\left(x-x_1\right) \\ \Rightarrow y-2=1\left(x-1\right) \\ \Rightarrow y-2=x-1 \\ \Rightarrow x-y+1=0 \end{gathered}$$

$$\begin{gathered} \text{Equation of normal is,} \\ y-y_1=\frac{-1}{m}\left(x-x_1\right) \\ \Rightarrow y-2=-1\left(x-1\right) \\ \Rightarrow y-2=-x+1 \\ \Rightarrow x+y-3=0 \end{gathered}$$

(xvii) Equation of tangent:
$$\begin{gathered} 4x^2+9y^2=36 \\ \text{Differentiating both sides w.r.t.}x, \\ 8x+18y\frac{dy}{dx}=0 \\ \Rightarrow 18y\frac{dy}{dx}=-8x \\ \Rightarrow \frac{dy}{dx}=\frac{-8x}{18y}=\frac{-4x}{9y} \\ \text{Slope of tangent,}\mathit{\text{m}}\text{=}{\left(\frac{dy}{dx}\right)}_{\left(3\cos \theta ,2\sin \theta \right)}\text{=}\frac{-12\cos \theta }{18\sin \theta }\text{=}\frac{-2\cos \theta }{3\sin \theta } \\ \mathrm{Given}\left(x_1,y_1\right)=\left(3\cos \theta ,2\sin \theta \right) \\ \text{Equation of tangent is,} \\ y-y_1=m\left(x-x_1\right) \\ \Rightarrow y-2\sin \theta =\frac{-2\cos \theta }{3\sin \theta }\left(x-3\cos \theta \right) \\ \Rightarrow 3y\sin \theta -6{\sin }^2\theta =-2x\cos \theta +6{\cos }^2\theta \\ \Rightarrow 2x\cos \theta +3y\sin \theta =6\left({\cos }^2\theta +{\sin }^2\theta \right) \\ \Rightarrow 2x\cos \theta +3y\sin \theta =6 \end{gathered}$$

$$\begin{gathered} \text{Equation of normal is,} \\ y-y_1=\frac{-1}{m}\left(x-x_1\right) \\ \Rightarrow y-2\sin \theta =\frac{3\sin \theta }{2\cos \theta }\left(x-3\cos \theta \right) \\ \Rightarrow 2y\cos \theta -4\sin \theta \cos \theta =3x\sin \theta -9\sin \theta \cos \theta \\ \Rightarrow 3x\sin \theta -2y\cos \theta -5\sin \theta \cos \theta =0 \end{gathered}$$

(xviii)
$$\begin{gathered} {\mathit{\text{y}}}^{\mathit{2}}\text{=4}\mathit{\text{ax}} \\ \text{Differentiating both sides w.r.t.}x, \\ 2y\frac{dy}{dx}=4a \\ \Rightarrow \frac{dy}{dx}=\frac{2a}{y} \\ \mathrm{At}\left(x_1,y_1\right) \\ \text{Slope of tangent =}{\left(\frac{dy}{dx}\right)}_{\left(x_1,y_1\right)}\text{=}\frac{2a}{{\text{y}}_1}\text{=}\mathit{\text{m}} \\ \text{Equation of tangent is,} \\ y-y_1=m\left(x-x_1\right) \\ \Rightarrow y-y_1=\frac{2a\left(x-x_1\right)}{y_1} \\ \Rightarrow y{y}_1-{y_1}^2=2ax-2a{x}_1 \\ \Rightarrow y{y}_1-4a{x}_1=2ax-2a{x}_1 \\ \Rightarrow y{y}_1=2ax+2a{x}_1 \\ \Rightarrow y{y}_1=2a\left(x+x_1\right) \end{gathered}$$

$$\begin{gathered} \text{Equation of normal is,} \\ y-y_1=\frac{1}{\mathrm{Slope}\mathrm{of}\mathrm{tangent}}\left(x-x_1\right) \\ \Rightarrow y-y_1=\frac{-1}{m}\left(x-x_1\right) \\ \Rightarrow y-y_1=\frac{-y_1}{2a}\left(x-x_1\right) \end{gathered}$$

(xix)
$$\begin{gathered} \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \\ \text{Differentiating both sides w.r.t.}x, \\ \frac{2x}{a^2}-\frac{2y}{b^2}\frac{dy}{dx}=0 \\ \Rightarrow \frac{2y}{b^2}\frac{dy}{dx}=\frac{2x}{a^2} \\ \Rightarrow \frac{dy}{dx}=\frac{x{b}^2}{y{a}^2} \\ \text{Slope of tangent,}\mathit{\text{m}}\text{=}{\left(\frac{dy}{dx}\right)}_{\left(\sqrt{2}\text{a,b}\right)}\text{=}\frac{\sqrt{2}\text{a}{b}^2}{b{a}^2}\text{=}\frac{\sqrt{2}\text{b}}{a} \\ \text{Equation of tangent is,} \\ y-y_1=m\left(x-x_1\right) \\ \Rightarrow y-b=\frac{\sqrt{2}\text{b}}{a}\left(x-\sqrt{2}a\right) \\ \Rightarrow ay-ab=\sqrt{2}bx-2ab \\ \Rightarrow \sqrt{2}bx-ay=ab \\ \Rightarrow \frac{\sqrt{2}x}{a}-\frac{y}{b}=1 \end{gathered}$$

$$\begin{gathered} \text{Equation of normal is,} \\ y-y_1=\frac{-1}{m}\left(x-x_1\right) \\ \Rightarrow y-b=\frac{-a}{\sqrt{2}b}\left(x-\sqrt{2}a\right) \\ \Rightarrow \sqrt{2}by-\sqrt{2}{b}^2=-ax+\sqrt{2}{a}^2 \\ \Rightarrow ax+\sqrt{2}by=\sqrt{2}{b}^2+\sqrt{2}{a}^2 \\ \Rightarrow \frac{ax}{\sqrt{2}}+by=a^2+b^2 \end{gathered}$$