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


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

$If x^{2} + y^{2} = 4, find minimum and maximum$ $value of \frac{x}{y + 3} .$
	Answered by liberty last updated on 22/Jan/21

	Answered by benjo_mathlover last updated on 22/Jan/21
	$let (x/(y+3)) = (1/m); m≠0 ⇒y+3 = mx or y = mx−3 substitution to circle give x^2 +(mx−3)^2 −4=0 ; (1+m^2 )x^2 −6mx+5=0 put discriminant = 0 ; 36m^2 −4(1+m^2 )(5)=0 36m^2 −20m^2 −20=0 ; m^2 = ((20)/(16)) m = ± ((√5)/2) so we get (x/(y+3)) = ± (2/( (√5)))=±((2(√5))/5) → { ((max = ((2(√5))/5))),((min = −((2(√5))/5))) :}$
	$let \frac{x}{y + 3} = \frac{1}{m}; m \neq 0 \Rightarrow y + 3 = mx$ $or y = mx - 3 substitution to circle give$ $x^{2} + {(m x - 3)}^{2} - 4 = 0; (1 + m^{2}) x^{2} - 6 m x + 5 = 0$ $p u t d i s c r i m i n a n t = 0; {36 m}^{2} - 4 (1 + m^{2}) (5) = 0$ ${36 m}^{2} - {20 m}^{2} - 20 = 0; m^{2} = \frac{20}{16}$ $m = \pm \frac{\sqrt{5}}{2} so we get \frac{x}{y + 3} = \pm \frac{2}{\sqrt{5}} = \pm \frac{2 \sqrt{5}}{5}$ $\to {\begin{cases} \max = \frac{2 \sqrt{5}}{5} \\ \min = - \frac{2 \sqrt{5}}{5} \end{cases}$
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