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Question Number 129462 by bemath last updated on 16/Jan/21

If x and y satisfy of equation (x−1)^2 +(y+1)^2 =5 then maximum value of (x/(y+1)) equal to ?

$$\mathrm{If}\:\mathrm{x}\:\mathrm{and}\:\mathrm{y}\:\mathrm{satisfy}\:\mathrm{of}\:\mathrm{equation}\: \\ $$$$\:\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{2}} +\left(\mathrm{y}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{5}\:\mathrm{then}\:\mathrm{maximum} \\ $$$$\mathrm{value}\:\mathrm{of}\:\frac{\mathrm{x}}{\mathrm{y}+\mathrm{1}}\:\mathrm{equal}\:\mathrm{to}\:?\: \\ $$

Commented by bramlexs22 last updated on 16/Jan/21

let (x/(y+1))=m ⇒y+1=(x/m) then x^2 −2x+1+(x^2 /m^2 )−5=0 (1+(1/m^2 ))x^2 −2x−4=0 we have 1 solution for Δ=0 4−4(((1+m^2 )/m^2 ))(−4)=0 1+4(((1+m^2 )/m^2 ))=0 nothing solution for m∈ R

$$\mathrm{let}\:\frac{\mathrm{x}}{\mathrm{y}+\mathrm{1}}=\mathrm{m}\:\Rightarrow\mathrm{y}+\mathrm{1}=\frac{\mathrm{x}}{\mathrm{m}} \\ $$$$\mathrm{then}\:\mathrm{x}^{\mathrm{2}} −\mathrm{2x}+\mathrm{1}+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{m}^{\mathrm{2}} }−\mathrm{5}=\mathrm{0} \\ $$$$\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{m}^{\mathrm{2}} }\right)\mathrm{x}^{\mathrm{2}} −\mathrm{2x}−\mathrm{4}=\mathrm{0} \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{1}\:\mathrm{solution}\:\mathrm{for}\:\Delta=\mathrm{0} \\ $$$$\:\mathrm{4}−\mathrm{4}\left(\frac{\mathrm{1}+\mathrm{m}^{\mathrm{2}} }{\mathrm{m}^{\mathrm{2}} }\right)\left(−\mathrm{4}\right)=\mathrm{0}\: \\ $$$$\:\mathrm{1}+\mathrm{4}\left(\frac{\mathrm{1}+\mathrm{m}^{\mathrm{2}} }{\mathrm{m}^{\mathrm{2}} }\right)=\mathrm{0}\:\mathrm{nothing}\:\mathrm{solution} \\ $$$$\mathrm{for}\:\mathrm{m}\in\:\mathbb{R} \\ $$$$\: \\ $$

Commented by prakash jain last updated on 16/Jan/21

y=−1,x=±(√5)+1 (x/(y+1))→±∞ no minimum, no maxium

$${y}=−\mathrm{1},{x}=\pm\sqrt{\mathrm{5}}+\mathrm{1} \\ $$$$\frac{{x}}{{y}+\mathrm{1}}\rightarrow\pm\infty \\ $$$$\mathrm{no}\:\mathrm{minimum},\:\mathrm{no}\:\mathrm{maxium} \\ $$

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