If-x-2-y-2-4-find-minimum-and-maximum-value-of-x-y-3-

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 \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}