find-the-equation-of-the-tangent-and-normal-to-the-curve-xy-9-at-x-4-

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

f i n d t h e e q u a t i o n o f t h e t a n g e n t a n d

n o r m a l t o t h e c u r v e x y = 9 a t x = 4

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

slope the tangent line

\Rightarrow y + {xy}^{'} = 0 \Rightarrow y^{'} = - \frac{y}{x} = - \frac{(\frac{9}{4})}{4}

m = - \frac{9}{16} . tangent line :

y = - \frac{9}{16} (x - 4) + \frac{9}{4} .

normal line : y = \frac{16}{9} (x - 4) + \frac{9}{4}

Commented by gopikrishnan last updated on 27/Jan/20

t h a n k u s i r

Answered by MJS last updated on 27/Jan/20

y = f (x)

y^{'} = f^{'} (p) at x = p

tangent at x = p :

y = a x + b

a = f^{'} (p)

finding b :

f (p) = f^{'} (p) p + b \Rightarrow b = f (p) - f^{'} (p) p

\Rightarrow tangent at x = p :

y = f^{'} (p) x + f (p) - f^{'} (p) p

normal at x = p :

y = - \frac{1}{a} x + c

a = f^{'} (p)

finding c :

f (p) = - \frac{p}{f^{'} (p)} + c \Rightarrow c = f (p) + \frac{p}{f^{'} (p)}

\Rightarrow normal at x = p :

y = - \frac{x}{f^{'} (p)} + f (p) + \frac{p}{f^{'} (p)}

in our case f (x) = \frac{9}{x} \Rightarrow f^{'} (x) = - \frac{9}{x^{2}}

tangent at x = 4

y = - \frac{9}{16} x + \frac{9}{2}

normal at x = 4

y = \frac{16}{9} x - \frac{175}{36}

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