Given equation of tangent line of the curve y = (b/x^2 ) at point (x,y) is bx−4y=−21. The value of b =?


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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
	$gradient of tangent line ⇒ (b/4) = −((2b)/x^3 ) ⇒ { ((x^3 =−8 ; x=−2 ∧y = (b/4))),((⇒−2b−4((b/4))=−21 ; −3b=−21 ; b=7)) :}$
	$g r a d i e n t o f t a n g e n t l i n e \Rightarrow \frac{b}{4} = - \frac{2 b}{x^{3}}$ $\Rightarrow {\begin{cases} x^{3} = - 8; x = - 2 \land y = \frac{b}{4} \\ \Rightarrow - 2 b - 4 (\frac{b}{4}) = - 21; - 3 b = - 21; b = 7 \end{cases}$
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