determine the value of b for which y=((−x)/3) +b meets the graph of y^2 =x^3 orthogonally


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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

$\tan θ = \frac{y}{x} = \frac{d y}{d x} y = x \frac{d y}{d x}$ $3 = \frac{b - \frac{x}{3} + \frac{c}{3}}{x} [(1)] (m e m e t e n g e n t e)$ $b - \frac{x_{0}}{3} = \sqrt{x_{0}^{3}} (2) (i n t e r s e c t)$ $(1) \Rightarrow \frac{10}{3} x_{0} = b + \frac{c}{3}$ $x_{0} = \frac{10}{3} b + \frac{10 c}{9}$ $b = \frac{10}{3} x_{0} - \frac{c}{3} \Rightarrow$ $b - \frac{x_{0}}{3} = 3 x_{0} - \frac{c}{3} = f (x_{0})$ $o u 3 x_{0} + (b - \frac{10 x_{0}}{3}) = t a n g e n t e$ $1) - b - \frac{x_{0}}{3} + \frac{c}{3} = 3 x_{0}$ $2) - 3 x_{0} - \frac{c}{3} = \sqrt{x_{0}^{3}}$ $x^{3} - 9 x^{2} + 2 c x - {(\frac{c}{3})}^{2} = 0$ $(a c o m p l e t e r . . .)$ $A s u i v r e . . . . . . . . . . . . . . . . . . .$
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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