determine-the-value-of-b-for-which-y-x-3-b-meets-the-graph-of-y-2-x-3-orthogonally-

Question Number 188083 by Kalebwizeman last updated on 25/Feb/23

d e t e r m i n e t h e v a l u e o f b f o r w h i c h

y = \frac{- x}{3} + b m e e t s t h e g r a p h o f

y^{2} = x^{3} o r t h o g o n a l l y

Commented by a.lgnaoui last updated on 27/Feb/23

\tan θ = \frac{y}{x} = \frac{d y}{d x} y = x \frac{d y}{d x}

3 = \frac{b - \frac{x}{3} + \frac{c}{3}}{x} [(1)] (m e m e t e n g e n t e)

b - \frac{x_{0}}{3} = \sqrt{x_{0}^{3}} (2) (i n t e r s e c t)

(1) \Rightarrow \frac{10}{3} x_{0} = b + \frac{c}{3}

x_{0} = \frac{10}{3} b + \frac{10 c}{9}

b = \frac{10}{3} x_{0} - \frac{c}{3} \Rightarrow

b - \frac{x_{0}}{3} = 3 x_{0} - \frac{c}{3} = f (x_{0})

o u 3 x_{0} + (b - \frac{10 x_{0}}{3}) = t a n g e n t e

1) - b - \frac{x_{0}}{3} + \frac{c}{3} = 3 x_{0}

2) - 3 x_{0} - \frac{c}{3} = \sqrt{x_{0}^{3}}

x^{3} - 9 x^{2} + 2 c x - {(\frac{c}{3})}^{2} = 0

(a c o m p l e t e r \dots)

A s u i v r e \dots \dots \dots \dots \dots \dots .

Commented by a.lgnaoui last updated on 26/Feb/23

Commented by Kalebwizeman last updated on 26/Feb/23

t h a n k y o u

w h a t i s n o w t h e v a l u e o f b ?

Answered by mr W last updated on 27/Feb/23

a t i n t e r s e c t i o n p o i n t :

2 y \frac{d y}{d x} = 3 x^{2}

\frac{d y}{d x} = 3

\Rightarrow 2 y \times 3 = 3 x^{2}

\Rightarrow y = \frac{x^{2}}{2} > 0

o n t h e o t h e r s i d e

y^{2} = x^{3}

{(\frac{x^{2}}{2})}^{2} = x^{3}

\Rightarrow x = 4

\Rightarrow y = \frac{4^{2}}{2} = 8

8 = - \frac{4}{3} + b

\Rightarrow b = \frac{4}{3} + 8 = \frac{28}{3} ✓

Commented by mr W last updated on 27/Feb/23

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