Find-the-area-of-the-triangle-in-the-first-quadrant-between-the-x-and-y-axis-and-the-tangent-to-the-relationship-curve-y-5-x-x-5-and-x-dont-equal-0-at-0-5-

Question Number 179458 by cortano1 last updated on 29/Oct/22

Find the area of the triangle in the

first quadrant between the x and y

axis and the tangent to the

relationship curve y = \frac{5}{x} - \frac{x}{5} and x dont

equal 0 at (0, 5)

Commented by Acem last updated on 29/Oct/22

I s t h e l a s t l i n e i m p l y i n g t h a t t h e C_{f} {d o e s n}^{'} t

i n t e r s e c t w i t h t h e a x i s y^{'} y “ w e k n o w {t h a t}^{″} o r t h a t

p o i n t o f t h e t a n g e n t . S o {w h a t}^{'} s m e a n t b y t h e

“ a n d x \neq 0 a t {(0, 5)}^{″}

Commented by cortano1 last updated on 30/Oct/22

in this book like that

Answered by DvMc last updated on 29/Oct/22

W e n e e d a n l i n e a l e c u a t i o n f (x) o f

s l o p e m = y^{'} a n d t h a t c o n t a i n s t h e

p o i n t P (x, \frac{5}{x} - \frac{x}{5})

y^{'} = - \frac{5}{x^{2}} - \frac{1}{5}

t h e n

\frac{5}{x} - \frac{x}{5} = (- \frac{5}{x^{2}} - \frac{1}{5}) x + b

\frac{5}{x} - \frac{x}{5} = - \frac{5}{x} - \frac{x}{5} + b

\frac{10}{x} = b

N e x t

f (x) = (- \frac{5}{k^{2}} - \frac{1}{5}) x + \frac{10}{k}

W h e r e k i s t h e x o f t h e o r i g i n a l f u n c t i o n

i n w h i c h t h e t a n e n c y i s n e e d e d

T h e h e i g h t o f t r i a n g l e i s \frac{10}{k}

a n d i t s b a s e i s x w h e n f (x) = 0 :

0 = (- \frac{5}{k^{2}} - \frac{1}{5}) x + \frac{10}{k}

x = \frac{10 / k}{(25 + k^{2}) / 5 k^{2}} = \frac{50 k}{25 + k^{2}}

S o, t h e a r e a o f t r i a n g l e t a n g e n t e t o y

i n x = k i s

A = \frac{1}{2} \times \frac{10}{k} \times \frac{50 k}{25 + k^{2}} = \frac{250}{25 + k^{2}}

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