Math Question and Answers


Question and Answers Forum

All Questions Topic List
Differentiation Questions
Previous in All Question Next in All Question
Previous in Differentiation Next in Differentiation
Question Number 39747 by Raj Singh last updated on 10/Jul/18

	Answered by tanmay.chaudhury50@gmail.com last updated on 10/Jul/18

	$y = \frac{x^{2}}{2 a - x} \frac{d y}{d x} = \frac{(2 a - x) 2 x - x^{2} (0 - 1)}{{(2 a - x)}^{2}}$ $\frac{d y}{d x} = \frac{4 a x - 2 x^{2} + x^{2}}{{(2 a - x)}^{2}} = \frac{4 a x - x^{2}}{{(2 a - x)}^{2}}$ $s l o p e o f t a n g e n t m = \frac{4 a^{2} - a^{2}}{a^{2}} = 3$ $e q n t a n g e n t (y - a) = 3 (x - a)$ $e q n n o r m a l y - a = \frac{- 1}{3} (x - a)$ $t a n g e n t c u t x a x i s a t (\frac{2 a}{3}, 0)$ $n o r m a l c u t x a x i s (\frac{4 a}{}, 0)$ $t r i a n g l e f o r m e d b y (a, a), (4 a, 0), (\frac{2 a}{3}, 0)$ $a r e a = \frac{1}{2} \times (4 a - \frac{2 a}{3}) \times a$ $= \frac{5 a^{2}}{3} [p l s c h e c k . . .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum