A curve has equation y = (2−(√x) )^4 The normal at point P(1,1) and the normal at point Q(9,1) intersect at the point R. What are coordinates at point R?


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 128521 by bramlexs22 last updated on 08/Jan/21

$A curve has equation y = {(2 - \sqrt{x})}^{4}$ $The normal at point P (1, 1) and$ $the normal at point Q (9, 1) intersect$ $at the point R . What are coordinates$ $at point R ?$
	Answered by mr W last updated on 08/Jan/21

	$\frac{d y}{d x} = 4 {(2 - \sqrt{x})}^{3} (- \frac{1}{2 \sqrt{x}}) = - \frac{2 {(2 - \sqrt{x})}^{3}}{\sqrt{x}}$ $a t P (1, 1) :$ $\frac{d y}{d x} = - 2$ $e q n . o f n o r m a l :$ $y = 1 + \frac{1}{2} (x - 1) = \frac{x + 1}{2}$ $a t Q (9, 1)$ $\frac{d y}{d x} = \frac{2}{3}$ $e q n . o f n o r m a l :$ $y = 1 - \frac{3}{2} (x - 9) = \frac{- 3 x + 29}{2}$ $i n t e r s e c t i o n :$ $y = \frac{x + 1}{2} = \frac{- 3 x + 29}{2}$ $\Rightarrow x = 7$ $\Rightarrow y = 4$ $\Rightarrow R (7, 4)$
	Commented by mr W last updated on 08/Jan/21

	Commented by bramlexs22 last updated on 08/Jan/21

	$great sir . thank you$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum