If-x-and-y-satisfy-of-equation-x-1-2-y-1-2-5-then-maximum-value-of-x-y-1-equal-to-

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 \frac{x}{y + 1} equal to ?

Commented by bramlexs22 last updated on 16/Jan/21

let \frac{x}{y + 1} = m \Rightarrow y + 1 = \frac{x}{m}

then x^{2} - 2 x + 1 + \frac{x^{2}}{m^{2}} - 5 = 0

(1 + \frac{1}{m^{2}}) x^{2} - 2 x - 4 = 0

we have 1 solution for Δ = 0

4 - 4 (\frac{1 + m^{2}}{m^{2}}) (- 4) = 0

1 + 4 (\frac{1 + m^{2}}{m^{2}}) = 0 nothing solution

for m \in R

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

y = - 1, x = \pm \sqrt{5} + 1

\frac{x}{y + 1} \to \pm \infty

no minimum, no maxium