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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).

$$\mathrm{Tangent}\:\mathrm{to}\:\mathrm{the}\:\mathrm{curve}\:\left(\mathrm{x}+\mathrm{y}\right)^{\mathrm{3}} =\left(\mathrm{x}−\mathrm{y}+\mathrm{2}\right)^{\mathrm{2}} \\ $$$$\mathrm{at}\:\left(−\mathrm{1},\mathrm{1}\right). \\ $$

Answered by ajfour last updated on 15/Mar/17

y=x+2

$${y}={x}+\mathrm{2} \\ $$

Commented by ajfour last updated on 15/Mar/17

Commented by ajfour last updated on 15/Mar/17

as (x+y)^3 =(x−y+2)^2   3(x+y)^2 (1+(dy/dx))=2(x−y+2)(1−(dy/dx))  let x=−1+h and y=1+k  and it should be noticed that  ((dy/dx))_(x=−1) =(k/h) = m (say)  ⇒ k = mh  the differentiated eqn now becomes  3(−1+h+mh+1)^2 (1+m)       = 2(−1+h−1−mh+2)(1−m)  3h^2 (1+m)^3  = 2h(1−m)^2   ⇒ as h→0  ,  m→ 1  so eqn of tangent is  y−1=m(x+1) and with m=1  it becomes  y=x+2 .

$${as}\:\left({x}+{y}\right)^{\mathrm{3}} =\left({x}−{y}+\mathrm{2}\right)^{\mathrm{2}} \\ $$$$\mathrm{3}\left({x}+{y}\right)^{\mathrm{2}} \left(\mathrm{1}+\frac{{dy}}{{dx}}\right)=\mathrm{2}\left({x}−{y}+\mathrm{2}\right)\left(\mathrm{1}−\frac{{dy}}{{dx}}\right) \\ $$$${let}\:{x}=−\mathrm{1}+{h}\:{and}\:{y}=\mathrm{1}+{k} \\ $$$${and}\:{it}\:{should}\:{be}\:{noticed}\:{that} \\ $$$$\left(\frac{{dy}}{{dx}}\right)_{{x}=−\mathrm{1}} =\frac{{k}}{{h}}\:=\:{m}\:\left({say}\right) \\ $$$$\Rightarrow\:{k}\:=\:{mh} \\ $$$${the}\:{differentiated}\:{eqn}\:{now}\:{becomes} \\ $$$$\mathrm{3}\left(−\mathrm{1}+{h}+{mh}+\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{1}+{m}\right) \\ $$$$\:\:\:\:\:=\:\mathrm{2}\left(−\mathrm{1}+{h}−\mathrm{1}−{mh}+\mathrm{2}\right)\left(\mathrm{1}−{m}\right) \\ $$$$\mathrm{3}{h}^{\mathrm{2}} \left(\mathrm{1}+{m}\right)^{\mathrm{3}} \:=\:\mathrm{2}{h}\left(\mathrm{1}−{m}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:{as}\:{h}\rightarrow\mathrm{0}\:\:,\:\:{m}\rightarrow\:\mathrm{1} \\ $$$${so}\:{eqn}\:{of}\:{tangent}\:{is} \\ $$$${y}−\mathrm{1}={m}\left({x}+\mathrm{1}\right)\:{and}\:{with}\:{m}=\mathrm{1} \\ $$$${it}\:{becomes}\:\:{y}={x}+\mathrm{2}\:. \\ $$

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