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Question Number 144534 by imjagoll last updated on 26/Jun/21

Find the shortest distance from the origin to the hyperbola x^2 +8xy+7y^2 =225 ,z=0

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{shortest}\:\mathrm{distance}\:\mathrm{from}\: \\ $$$$\mathrm{the}\:\mathrm{origin}\:\mathrm{to}\:\mathrm{the}\:\mathrm{hyperbola}\: \\ $$$$\mathrm{x}^{\mathrm{2}} +\mathrm{8xy}+\mathrm{7y}^{\mathrm{2}} =\mathrm{225}\:,\mathrm{z}=\mathrm{0}\: \\ $$

Answered by liberty last updated on 26/Jun/21

we must find the minimum value of x^2 +y^2 (the square of the distance from the origin to any point in the xy plane) subject to constraint x^2 +8xy+7y^2 =225 by Langrange multiplier ∅=x^2 +8xy+7y^2 +λ(x^2 +y^2 ) ∅_x =2x+8y+2λx=0 or (λ+1)x+4y=0 φ_y =8x+14y+2λy=0 or 4x+(λ+7)y=0 determinant (((λ+1 4)),(( 4 λ+7)))=0 → { ((λ=−9)),((λ=1)) :} when λ=−9 we get y=2x and substitution in x^2 +8xy+7y^2 =225 yields x^2 =5, y^2 =20 so the shortest distance is (√(5+20))=5 when λ=1 give x=−2y and substitution in x^2 +8xy+7y^2 =225 yields −5y^2 =225 for which no real solution exists.

$$\mathrm{we}\:\mathrm{must}\:\mathrm{find}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{value} \\ $$$$\mathrm{of}\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \:\left(\mathrm{the}\:\mathrm{square}\:\mathrm{of}\:\mathrm{the}\:\mathrm{distance}\right. \\ $$$$\mathrm{from}\:\mathrm{the}\:\mathrm{origin}\:\mathrm{to}\:\mathrm{any}\:\mathrm{point}\:\mathrm{in} \\ $$$$\left.\mathrm{the}\:\mathrm{xy}\:\mathrm{plane}\right)\:\mathrm{subject}\:\mathrm{to}\:\mathrm{constraint} \\ $$$$\mathrm{x}^{\mathrm{2}} +\mathrm{8xy}+\mathrm{7y}^{\mathrm{2}} =\mathrm{225} \\ $$$$\mathrm{by}\:\mathrm{Langrange}\:\mathrm{multiplier} \\ $$$$\emptyset=\mathrm{x}^{\mathrm{2}} +\mathrm{8xy}+\mathrm{7y}^{\mathrm{2}} +\lambda\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \right) \\ $$$$\emptyset_{\mathrm{x}} =\mathrm{2x}+\mathrm{8y}+\mathrm{2}\lambda\mathrm{x}=\mathrm{0}\:\mathrm{or}\:\left(\lambda+\mathrm{1}\right)\mathrm{x}+\mathrm{4y}=\mathrm{0} \\ $$$$\phi_{\mathrm{y}} =\mathrm{8x}+\mathrm{14y}+\mathrm{2}\lambda\mathrm{y}=\mathrm{0}\:\mathrm{or}\:\mathrm{4x}+\left(\lambda+\mathrm{7}\right)\mathrm{y}=\mathrm{0} \\ $$$$\begin{vmatrix}{\lambda+\mathrm{1}\:\:\:\:\:\mathrm{4}}\\{\:\:\:\mathrm{4}\:\:\:\:\lambda+\mathrm{7}}\end{vmatrix}=\mathrm{0}\:\rightarrow\begin{cases}{\lambda=−\mathrm{9}}\\{\lambda=\mathrm{1}}\end{cases} \\ $$$$\mathrm{when}\:\lambda=−\mathrm{9}\:\mathrm{we}\:\mathrm{get}\:\mathrm{y}=\mathrm{2x}\:\mathrm{and}\: \\ $$$$\mathrm{substitution}\:\mathrm{in}\:\mathrm{x}^{\mathrm{2}} +\mathrm{8xy}+\mathrm{7y}^{\mathrm{2}} =\mathrm{225} \\ $$$$\mathrm{yields}\:\mathrm{x}^{\mathrm{2}} =\mathrm{5},\:\mathrm{y}^{\mathrm{2}} =\mathrm{20}\:\mathrm{so}\:\mathrm{the}\:\mathrm{shortest} \\ $$$$\mathrm{distance}\:\mathrm{is}\:\sqrt{\mathrm{5}+\mathrm{20}}=\mathrm{5} \\ $$$$\mathrm{when}\:\lambda=\mathrm{1}\:\mathrm{give}\:\mathrm{x}=−\mathrm{2y}\:\mathrm{and}\: \\ $$$$\mathrm{substitution}\:\mathrm{in}\:\mathrm{x}^{\mathrm{2}} +\mathrm{8xy}+\mathrm{7y}^{\mathrm{2}} =\mathrm{225} \\ $$$$\mathrm{yields}\:−\mathrm{5y}^{\mathrm{2}} =\mathrm{225}\:\mathrm{for}\:\mathrm{which}\:\mathrm{no} \\ $$$$\mathrm{real}\:\mathrm{solution}\:\mathrm{exists}. \\ $$

Answered by mr W last updated on 26/Jun/21

r=distance from (0,0) to (x,y) x=r cos θ y=r sin θ r^2 cos^2 θ+8r^2 cos θ sin θ+7r^2 sin^2 θ=225 r^2 (4+4sin 2θ−3cos 2θ)=225 r^2 =((225)/(4+5 sin (2θ−tan^(−1) (3/4))))≥((225)/(4+5)) ⇒r_(min) =(√((225)/9))=5

$${r}={distance}\:{from}\:\left(\mathrm{0},\mathrm{0}\right)\:{to}\:\left({x},{y}\right) \\ $$$${x}={r}\:\mathrm{cos}\:\theta \\ $$$${y}={r}\:\mathrm{sin}\:\theta \\ $$$${r}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\theta+\mathrm{8}{r}^{\mathrm{2}} \mathrm{cos}\:\theta\:\mathrm{sin}\:\theta+\mathrm{7}{r}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta=\mathrm{225} \\ $$$${r}^{\mathrm{2}} \left(\mathrm{4}+\mathrm{4sin}\:\mathrm{2}\theta−\mathrm{3cos}\:\mathrm{2}\theta\right)=\mathrm{225} \\ $$$${r}^{\mathrm{2}} =\frac{\mathrm{225}}{\mathrm{4}+\mathrm{5}\:\mathrm{sin}\:\left(\mathrm{2}\theta−\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{3}}{\mathrm{4}}\right)}\geqslant\frac{\mathrm{225}}{\mathrm{4}+\mathrm{5}} \\ $$$$\Rightarrow{r}_{{min}} =\sqrt{\frac{\mathrm{225}}{\mathrm{9}}}=\mathrm{5} \\ $$

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