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Question Number 79667 by gopikrishnan last updated on 27/Jan/20

$$findtheequationofthetangentand$$$$normaltothecurvexy=9atx=4$$

Commented by john santu last updated on 27/Jan/20

$$\mathrm{slope}\mathrm{the}\mathrm{tangent}\mathrm{line}$$$$\Rightarrow \mathrm{y}+{\mathrm{xy}}^{\prime }=0\Rightarrow {\mathrm{y}}^{\prime }=-\frac{\mathrm{y}}{\mathrm{x}}=-\frac{\left(\frac{9}{4}\right)}{4}$$$$\mathrm{m}=-\frac{9}{16}.\mathrm{tangent}\mathrm{line}:$$$$\mathrm{y}=-\frac{9}{16}(\mathrm{x}-4)+\frac{9}{4}.$$$$\mathrm{normal}\mathrm{line}:\mathrm{y}=\frac{16}{9}(\mathrm{x}-4)+\frac{9}{4}$$

Commented by gopikrishnan last updated on 27/Jan/20

$$thankusir$$

Answered by MJS last updated on 27/Jan/20

$$y=f\left(x\right)$$$$y^{\prime }=f^{\prime }\left(p\right)\mathrm{at}x=p$$$$\mathrm{tangent}\mathrm{at}x=p:$$$$y=ax+b$$$$a=f^{\prime }\left(p\right)$$$$\mathrm{finding}b:$$$$f\left(p\right)=f^{\prime }\left(p\right)p+b\Rightarrow b=f\left(p\right)-f^{\prime }\left(p\right)p$$$$\Rightarrow \mathrm{tangent}\mathrm{at}x=p:$$$$y=f^{\prime }\left(p\right)x+f\left(p\right)-f^{\prime }\left(p\right)p$$$$$$$$\mathrm{normal}\mathrm{at}x=p:$$$$y=-\frac{1}{a}x+c$$$$a=f^{\prime }\left(p\right)$$$$\mathrm{finding}c:$$$$f\left(p\right)=-\frac{p}{f^{\prime }\left(p\right)}+c\Rightarrow c=f\left(p\right)+\frac{p}{f^{\prime }\left(p\right)}$$$$\Rightarrow \mathrm{normal}\mathrm{at}x=p:$$$$y=-\frac{x}{f^{\prime }\left(p\right)}+f\left(p\right)+\frac{p}{f^{\prime }\left(p\right)}$$$$$$$$\mathrm{in}\mathrm{our}\mathrm{case}f\left(x\right)=\frac{9}{x}\Rightarrow f^{\prime }\left(x\right)=-\frac{9}{x^2}$$$$\mathrm{tangent}\mathrm{at}x=4$$$$y=-\frac{9}{16}x+\frac{9}{2}$$$$\mathrm{normal}\mathrm{at}x=4$$$$y=\frac{16}{9}x-\frac{175}{36}$$

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