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Question Number 130077 by benjo_mathlover last updated on 22/Jan/21

If x^2 +y^2 =4 , find minimum and maximum value of (x/(y+3)).

$$\:\mathrm{If}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{4}\:,\:\mathrm{find}\:\mathrm{minimum}\:\mathrm{and}\:\mathrm{maximum} \\ $$$$\mathrm{value}\:\mathrm{of}\:\frac{\mathrm{x}}{\mathrm{y}+\mathrm{3}}. \\ $$

Answered by liberty last updated on 22/Jan/21

Answered by benjo_mathlover last updated on 22/Jan/21

let (x/(y+3)) = (1/m); m≠0 ⇒y+3 = mx or y = mx−3 substitution to circle give x^2 +(mx−3)^2 −4=0 ; (1+m^2 )x^2 −6mx+5=0 put discriminant = 0 ; 36m^2 −4(1+m^2 )(5)=0 36m^2 −20m^2 −20=0 ; m^2 = ((20)/(16)) m = ± ((√5)/2) so we get (x/(y+3)) = ± (2/( (√5)))=±((2(√5))/5) → { ((max = ((2(√5))/5))),((min = −((2(√5))/5))) :}

$$\:\mathrm{let}\:\frac{\mathrm{x}}{\mathrm{y}+\mathrm{3}}\:=\:\frac{\mathrm{1}}{\mathrm{m}};\:\mathrm{m}\neq\mathrm{0}\:\Rightarrow\mathrm{y}+\mathrm{3}\:=\:\mathrm{mx}\: \\ $$$$\mathrm{or}\:\mathrm{y}\:=\:\mathrm{mx}−\mathrm{3}\:\mathrm{substitution}\:\mathrm{to}\:\mathrm{circle}\:\mathrm{give}\: \\ $$$$\:{x}^{\mathrm{2}} +\left({mx}−\mathrm{3}\right)^{\mathrm{2}} −\mathrm{4}=\mathrm{0}\:;\:\left(\mathrm{1}+\mathrm{m}^{\mathrm{2}} \right){x}^{\mathrm{2}} −\mathrm{6}{mx}+\mathrm{5}=\mathrm{0} \\ $$$${put}\:{discriminant}\:=\:\mathrm{0}\:;\:\mathrm{36m}^{\mathrm{2}} −\mathrm{4}\left(\mathrm{1}+\mathrm{m}^{\mathrm{2}} \right)\left(\mathrm{5}\right)=\mathrm{0} \\ $$$$\:\mathrm{36m}^{\mathrm{2}} −\mathrm{20m}^{\mathrm{2}} −\mathrm{20}=\mathrm{0}\:;\:\mathrm{m}^{\mathrm{2}} \:=\:\frac{\mathrm{20}}{\mathrm{16}} \\ $$$$\:\mathrm{m}\:=\:\pm\:\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\:\mathrm{so}\:\mathrm{we}\:\mathrm{get}\:\frac{\mathrm{x}}{\mathrm{y}+\mathrm{3}}\:=\:\pm\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}=\pm\frac{\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{5}} \\ $$$$\:\rightarrow\begin{cases}{\mathrm{max}\:=\:\frac{\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{5}}}\\{\mathrm{min}\:=\:−\frac{\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{5}}}\end{cases} \\ $$

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