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 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