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Question Number 161166 by naka3546 last updated on 13/Dec/21

$$If{}^4\log (x+2y)+{}^4\log (x-2y)=1.$$$$Minimumvalueof\mid x\mid -\mid y\mid is...?$$

Answered by mr W last updated on 13/Dec/21

$$(x+2y)(x-2y)=4$$$${\left(\frac{x}{2}\right)}^2-y^2=1\Rightarrow hyperbola$$$$2\left(\frac{x}{2}\right)\frac{1}{2}-2y\frac{dy}{dx}=0$$$$2\left(\frac{x}{2}\right)\frac{1}{2}-2y\times 1=0$$$$y=\frac{x}{4}$$$${\left(\frac{x}{2}\right)}^2-{\left(\frac{x}{4}\right)}^2=1$$$$\Rightarrow x=\pm \frac{4}{\sqrt{3}}$$$$\Rightarrow y=\pm \frac{1}{\sqrt{3}}$$$$\mid x\mid -\mid y\mid =\frac{4}{\sqrt{3}}-\frac{1}{\sqrt{3}}=\sqrt{3}=minimun$$

Commented by naka3546 last updated on 13/Dec/21

$$Thankyou,sir.$$

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