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Question Number 140288 by liberty last updated on 06/May/21

Find the point(s) on the graph of 3x^2 +10xy+3y^2 =9 closest to the origin

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{point}\left(\mathrm{s}\right)\:\mathrm{on}\:\mathrm{the}\:\mathrm{graph} \\ $$$$\mathrm{of}\:\mathrm{3x}^{\mathrm{2}} +\mathrm{10xy}+\mathrm{3y}^{\mathrm{2}} =\mathrm{9}\:\mathrm{closest}\:\mathrm{to} \\ $$$$\mathrm{the}\:\mathrm{origin} \\ $$

Answered by mr W last updated on 06/May/21

with y=x: 3x^2 +10x^2 +3x^2 =9 16x^2 =9 x=±(3/4) ⇒y=±(3/4) with y=−x: 3x^2 −10x^2 +3x^2 =9 −x^2 =9 ⇒no solution point ((3/4),(3/4)) and (−(3/4),−(3/4)) are closest to origin.

$${with}\:{y}={x}: \\ $$$$\mathrm{3}{x}^{\mathrm{2}} +\mathrm{10}{x}^{\mathrm{2}} +\mathrm{3}{x}^{\mathrm{2}} =\mathrm{9} \\ $$$$\mathrm{16}{x}^{\mathrm{2}} =\mathrm{9} \\ $$$${x}=\pm\frac{\mathrm{3}}{\mathrm{4}}\:\Rightarrow{y}=\pm\frac{\mathrm{3}}{\mathrm{4}} \\ $$$${with}\:{y}=−{x}: \\ $$$$\mathrm{3}{x}^{\mathrm{2}} −\mathrm{10}{x}^{\mathrm{2}} +\mathrm{3}{x}^{\mathrm{2}} =\mathrm{9} \\ $$$$−{x}^{\mathrm{2}} =\mathrm{9} \\ $$$$\Rightarrow{no}\:{solution} \\ $$$$ \\ $$$${point}\:\left(\frac{\mathrm{3}}{\mathrm{4}},\frac{\mathrm{3}}{\mathrm{4}}\right)\:{and}\:\left(−\frac{\mathrm{3}}{\mathrm{4}},−\frac{\mathrm{3}}{\mathrm{4}}\right)\:{are} \\ $$$${closest}\:{to}\:{origin}. \\ $$

Commented by liberty last updated on 06/May/21

why y=x ?

$$\mathrm{why}\:\mathrm{y}=\mathrm{x}\:?\: \\ $$

Commented by mr W last updated on 06/May/21

symmetry of hyperbola

$${symmetry}\:{of}\:{hyperbola} \\ $$

Commented by liberty last updated on 06/May/21

Commented by liberty last updated on 06/May/21

oo do you meant like this

$$\mathrm{oo}\:\mathrm{do}\:\mathrm{you}\:\mathrm{meant}\:\mathrm{like}\:\mathrm{this} \\ $$

Commented by mr W last updated on 06/May/21

yes

$${yes} \\ $$

Answered by liberty last updated on 06/May/21

let(x,y) is a point on curve 3x^2 +10xy+3y^2 =9 closest to the origin. square the distance from the origin u = x^2 +y^2 . by implicit diff (1)(du/dx) = 2x+ 2y(dy/dx) (2) from the second eq 6x + 10y +10x (dy/dx) + 6y (dy/dx) = 0 we get (dy/dx) =−((3x+5y)/(5x+3y)) substitute to eq(1) ⇒ (du/dx) = 2x−2y(((3x+5y)/(5x+3y)))=0 ⇒x^2 =y^2 ; x = ± y substituting in the eq of the graph 16x^2 =9 → { ((x=±(3/4))),((y=± (3/4))) :}

$$\mathrm{let}\left(\mathrm{x},\mathrm{y}\right)\:\mathrm{is}\:\mathrm{a}\:\mathrm{point}\:\mathrm{on}\:\mathrm{curve}\:\mathrm{3x}^{\mathrm{2}} +\mathrm{10xy}+\mathrm{3y}^{\mathrm{2}} =\mathrm{9} \\ $$$$\mathrm{closest}\:\mathrm{to}\:\mathrm{the}\:\mathrm{origin}.\:\mathrm{square}\:\mathrm{the} \\ $$$$\mathrm{distance}\:\mathrm{from}\:\mathrm{the}\:\mathrm{origin} \\ $$$$\:\mathrm{u}\:=\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \:.\:\mathrm{by}\:\mathrm{implicit}\:\mathrm{diff} \\ $$$$\left(\mathrm{1}\right)\frac{\mathrm{du}}{\mathrm{dx}}\:=\:\mathrm{2x}+\:\mathrm{2y}\frac{\mathrm{dy}}{\mathrm{dx}} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{from}\:\mathrm{the}\:\mathrm{second}\:\mathrm{eq}\: \\ $$$$\:\mathrm{6x}\:+\:\mathrm{10y}\:+\mathrm{10x}\:\frac{\mathrm{dy}}{\mathrm{dx}}\:+\:\mathrm{6y}\:\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\mathrm{0} \\ $$$$\mathrm{we}\:\mathrm{get}\:\frac{\mathrm{dy}}{\mathrm{dx}}\:=−\frac{\mathrm{3x}+\mathrm{5y}}{\mathrm{5x}+\mathrm{3y}} \\ $$$$\mathrm{substitute}\:\mathrm{to}\:\mathrm{eq}\left(\mathrm{1}\right) \\ $$$$\Rightarrow\:\frac{\mathrm{du}}{\mathrm{dx}}\:=\:\mathrm{2x}−\mathrm{2y}\left(\frac{\mathrm{3x}+\mathrm{5y}}{\mathrm{5x}+\mathrm{3y}}\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{x}^{\mathrm{2}} =\mathrm{y}^{\mathrm{2}} ;\:\mathrm{x}\:=\:\pm\:\mathrm{y} \\ $$$$\mathrm{substituting}\:\mathrm{in}\:\mathrm{the}\:\mathrm{eq}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{graph}\:\mathrm{16x}^{\mathrm{2}} =\mathrm{9}\:\rightarrow\begin{cases}{\mathrm{x}=\pm\frac{\mathrm{3}}{\mathrm{4}}}\\{\mathrm{y}=\pm\:\frac{\mathrm{3}}{\mathrm{4}}}\end{cases} \\ $$

Commented by lyubita last updated on 06/May/21

tepebea

$${tepebea} \\ $$

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