Given-equation-of-tangent-line-of-the-curve-y-b-x-2-at-point-x-y-is-bx-4y-21-The-value-of-b-

Question Number 124133 by bramlexs22 last updated on 01/Dec/20

G i v e n e q u a t i o n o f t a n g e n t l i n e

o f t h e c u r v e y = \frac{b}{x^{2}} a t p o i n t (x, y)

i s b x - 4 y = - 21 . T h e v a l u e o f b = ?

Answered by mr W last updated on 01/Dec/20

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

t a n g e n t a t p o i n t (p, \frac{b}{p^{2}})

y = - \frac{2 b}{p^{3}} (x - p) + \frac{b}{p^{2}}

\Rightarrow b x + \frac{p^{3}}{2} y = \frac{3}{2} p b

\frac{p^{3}}{2} = - 4 \Rightarrow p = - 2

\frac{3}{2} \times (- 2) b = - 21 \Rightarrow b = 7

Commented by mr W last updated on 01/Dec/20

Commented by I want to learn more last updated on 01/Dec/20

Sir mrW, please help me with permutation on Q 124149

Answered by liberty last updated on 01/Dec/20

g r a d i e n t o f t a n g e n t l i n e \Rightarrow \frac{b}{4} = - \frac{2 b}{x^{3}}

\Rightarrow {\begin{cases} x^{3} = - 8; x = - 2 \land y = \frac{b}{4} \\ \Rightarrow - 2 b - 4 (\frac{b}{4}) = - 21; - 3 b = - 21; b = 7 \end{cases}

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