Tangent-to-the-curve-x-y-3-x-y-2-2-at-1-1-

Question Number 11172 by agni5 last updated on 15/Mar/17

Tangent to the curve {(x + y)}^{3} = {(x - y + 2)}^{2}

at (- 1, 1) .

Answered by ajfour last updated on 15/Mar/17

y = x + 2

Commented by ajfour last updated on 15/Mar/17

Commented by ajfour last updated on 15/Mar/17

a s {(x + y)}^{3} = {(x - y + 2)}^{2}

3 {(x + y)}^{2} (1 + \frac{d y}{d x}) = 2 (x - y + 2) (1 - \frac{d y}{d x})

l e t x = - 1 + h a n d y = 1 + k

a n d i t s h o u l d b e n o t i c e d t h a t

{(\frac{d y}{d x})}_{x = - 1} = \frac{k}{h} = m (s a y)

\Rightarrow k = m h

t h e d i f f e r e n t i a t e d e q n n o w b e c o m e s

3 {(- 1 + h + m h + 1)}^{2} (1 + m)

= 2 (- 1 + h - 1 - m h + 2) (1 - m)

3 h^{2} {(1 + m)}^{3} = 2 h {(1 - m)}^{2}

\Rightarrow a s h \to 0, m \to 1

s o e q n o f t a n g e n t i s

y - 1 = m (x + 1) a n d w i t h m = 1

i t b e c o m e s y = x + 2 .

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