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Question Number 120378 by john santu last updated on 31/Oct/20

$$findthepointonthecurve3x^2-4y^2=72$$$$closesttoline3x+2y-1=0$$

Commented by bramlexs22 last updated on 31/Oct/20

$$let(x,y)isapointonhyperbola$$$$y^2=\frac{3x^2-72}{4}andclosesttoline$$$$3x+2y-1=0,wegetthedistance$$$$d=\frac{\mid 3x\pm 2\sqrt{\frac{3x^2-72}{4}}-1\mid }{\sqrt{13}}$$$$d=\frac{\mid 3x\pm \sqrt{3x^2-72}-1\mid }{\sqrt{13}}$$$$wetake\frac{d}{dx}\left[\frac{3x\pm \sqrt{3x^2-72}-1}{\sqrt{13}}\right]=0$$$$\iff 3\pm \frac{3x}{\sqrt{3x^2-72}}=0$$$$weget{x}^2=36orx=\pm 6then$$$$y=\pm \sqrt{\frac{3.36-72}{4}}=\pm \sqrt{9}=\pm 3$$$$checkpoint$$$$\left(i\right)(6,3)\Rightarrow d=\frac{18+6-1}{\sqrt{13}}=\frac{23}{\sqrt{13}}$$$$\left(ii\right)(6,-3)\Rightarrow d=\frac{18-6-1}{\sqrt{13}}=\frac{11}{\sqrt{13}}\leftarrow d_{min}$$$$\left(iii\right)(-6,3)\Rightarrow d=\frac{\mid -18+6-1\mid }{\sqrt{13}}=\frac{17}{\sqrt{13}}$$$$\left(iv\right)(-6,-3)\Rightarrow d=\frac{\mid -18-6-1\mid }{\sqrt{13}}=\frac{25}{\sqrt{13}}$$

Commented by john santu last updated on 31/Oct/20

Answered by mr W last updated on 31/Oct/20

$$3x^2-4y^2=72$$$$6x-8y\frac{dy}{dx}=0$$$$6x-8y\times (-\frac{3}{2})=0$$$$\Rightarrow x=-2y$$$$3{(-2y)}^2-4y^2=72$$$$y^2=9$$$$\Rightarrow y=\pm 3$$$$\Rightarrow x=\mp 6$$$$\Rightarrow (6,-3)and(-6,3)$$$$\frac{1}{\sqrt{3^2+2^2}}\mid 3\times 6+2(-3)-1\mid =\frac{11}{\sqrt{13}}✓$$$$\frac{1}{\sqrt{3^2+2^2}}\mid 3\times (-6)+2\times 3-1\mid =\frac{13}{\sqrt{13}}$$$$\Rightarrow (6,-3)isclosestto3x+2y-1=0.$$$$andtheclosestdistanceis\frac{11}{\sqrt{13}}.$$

Commented by mr W last updated on 31/Oct/20

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