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Question Number 79667 by gopikrishnan last updated on 27/Jan/20
find the equation of the tangent and normal to the curve xy=9 at x=4
$${find}\:{the}\:{equation}\:{of}\:{the}\:{tangent}\:{and} \\ $$$${normal}\:{to}\:{the}\:{curve}\:{xy}=\mathrm{9}\:{at}\:{x}=\mathrm{4} \\ $$
Commented by john santu last updated on 27/Jan/20
slope the tangent line ⇒y+xy′ = 0 ⇒y′ =−(y/x)=−((((9/4)))/4) m= −(9/(16)) .tangent line : y= −(9/(16))(x−4)+(9/4). normal line : y = ((16)/9)(x−4)+(9/4)
$$\mathrm{slope}\:\mathrm{the}\:\mathrm{tangent}\:\mathrm{line} \\ $$$$\Rightarrow\mathrm{y}+\mathrm{xy}'\:=\:\mathrm{0}\:\Rightarrow\mathrm{y}'\:=−\frac{\mathrm{y}}{\mathrm{x}}=−\frac{\left(\frac{\mathrm{9}}{\mathrm{4}}\right)}{\mathrm{4}} \\ $$$$\mathrm{m}=\:−\frac{\mathrm{9}}{\mathrm{16}}\:.\mathrm{tangent}\:\mathrm{line}\::\: \\ $$$$\mathrm{y}=\:−\frac{\mathrm{9}}{\mathrm{16}}\left(\mathrm{x}−\mathrm{4}\right)+\frac{\mathrm{9}}{\mathrm{4}}. \\ $$$$\mathrm{normal}\:\mathrm{line}\::\:\mathrm{y}\:=\:\frac{\mathrm{16}}{\mathrm{9}}\left(\mathrm{x}−\mathrm{4}\right)+\frac{\mathrm{9}}{\mathrm{4}} \\ $$
Commented by gopikrishnan last updated on 27/Jan/20
thank u sir
$${thank}\:{u}\:{sir} \\ $$
Answered by MJS last updated on 27/Jan/20
y=f(x) y′=f′(p) at x=p tangent at x=p: y=ax+b a=f′(p) finding b: f(p)=f′(p)p+b ⇒ b=f(p)−f′(p)p ⇒ tangent at x=p: y=f′(p)x+f(p)−f′(p)p normal at x=p: y=−(1/a)x+c a=f′(p) finding c: f(p)=−(p/(f′(p)))+c ⇒ c=f(p)+(p/(f′(p))) ⇒ normal at x=p: y=−(x/(f′(p)))+f(p)+(p/(f′(p))) in our case f(x)=(9/x) ⇒ f′(x)=−(9/x^2 ) tangent at x=4 y=−(9/(16))x+(9/2) normal at x=4 y=((16)/9)x−((175)/(36))
$${y}={f}\left({x}\right) \\ $$$${y}'={f}'\left({p}\right)\:\mathrm{at}\:{x}={p} \\ $$$$\mathrm{tangent}\:\mathrm{at}\:{x}={p}: \\ $$$${y}={ax}+{b} \\ $$$${a}={f}'\left({p}\right) \\ $$$$\mathrm{finding}\:{b}: \\ $$$${f}\left({p}\right)={f}'\left({p}\right){p}+{b}\:\Rightarrow\:{b}={f}\left({p}\right)−{f}'\left({p}\right){p} \\ $$$$\Rightarrow\:\mathrm{tangent}\:\mathrm{at}\:{x}={p}: \\ $$$${y}={f}'\left({p}\right){x}+{f}\left({p}\right)−{f}'\left({p}\right){p} \\ $$$$ \\ $$$$\mathrm{normal}\:\mathrm{at}\:{x}={p}: \\ $$$${y}=−\frac{\mathrm{1}}{{a}}{x}+{c} \\ $$$${a}={f}'\left({p}\right) \\ $$$$\mathrm{finding}\:{c}: \\ $$$${f}\left({p}\right)=−\frac{{p}}{{f}'\left({p}\right)}+{c}\:\Rightarrow\:{c}={f}\left({p}\right)+\frac{{p}}{{f}'\left({p}\right)} \\ $$$$\Rightarrow\:\mathrm{normal}\:\mathrm{at}\:{x}={p}: \\ $$$${y}=−\frac{{x}}{{f}'\left({p}\right)}+{f}\left({p}\right)+\frac{{p}}{{f}'\left({p}\right)} \\ $$$$ \\ $$$$\mathrm{in}\:\mathrm{our}\:\mathrm{case}\:{f}\left({x}\right)=\frac{\mathrm{9}}{{x}}\:\Rightarrow\:{f}'\left({x}\right)=−\frac{\mathrm{9}}{{x}^{\mathrm{2}} } \\ $$$$\mathrm{tangent}\:\mathrm{at}\:{x}=\mathrm{4} \\ $$$${y}=−\frac{\mathrm{9}}{\mathrm{16}}{x}+\frac{\mathrm{9}}{\mathrm{2}} \\ $$$$\mathrm{normal}\:\mathrm{at}\:{x}=\mathrm{4} \\ $$$${y}=\frac{\mathrm{16}}{\mathrm{9}}{x}−\frac{\mathrm{175}}{\mathrm{36}} \\ $$

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