given-that-the-parametric-equation-of-a-curvd-are-x-1-t-1-y-1-t-1-obtaun-a-cartesian-equation-of-the-curve-Hemce-find-an-equqtion-of-tbe-nirmal-to-the-curve-at-the-point-t-2-

Question Number 103751 by hardylanes last updated on 17/Jul/20

g i v e n t h a t t h e p a r a m e t r i c e q u a t i o n o f

a c u r v d a r e x = \frac{1}{t - 1} y = \frac{1}{t + 1} o b t a u n a

c a r t e s i a n e q u a t i o n o f t h e c u r v e . H e m c e f i n d

a n e q u q t i o n o f t b e n i r m a l t o t h e c u r v e a t t h e

p o i n t t = 2

Answered by bobhans last updated on 17/Jul/20

x = \frac{1}{t - 1} \to t - 1 = \frac{1}{x}; t + 1 = \frac{1}{x} + 2

y = \frac{1}{t + 1} = \frac{x}{1 + 2 x} . \Rightarrow y^{'} = \frac{1}{{(2 x + 1)}^{2}}

s l o p e n o r m a l l i n e \Rightarrow m = - \frac{1}{y^{'}} = - {(2 x + 1)}^{2}

a t p o i n t (x, y) = (1, \frac{1}{3}) \Rightarrow m = - 9

e q o f t h e n o r m a l l i n e \Rightarrow 9 x + y = 9.1 + 1 . \frac{1}{3}

9 x + y = \frac{28}{3} o r 27 x + 3 y = 28 \oplus

Commented by bobhans last updated on 17/Jul/20

w h a t d o y o u m e a n t ? ? ?

Facebook Tweet Pin