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Question Number 186952 by Humble last updated on 12/Feb/23

Answered by horsebrand11 last updated on 12/Feb/23

let 3x+2y+c = 0 is tangent to hypebola . we have (1/(12))x−(1/9)y.y′=0 ⇒y′=(((3/(36))x)/((4/(36))y)) = ((3x)/(4y)) = −(3/2) ⇒x=−2y and ((4y^2 )/(24))−(y^2 /(18)) =1 ⇒y^2 =9 ⇒ { ((y=3 ; x=−6)),((y=−3 ; x=6)) :} So the contac point at (−6,3) or (6,−3) for (−6,3)⇒distance =((∣−18+6+1∣)/( (√(13))))=((11)/( (√(13)))) for (6,−3)⇒distance=((∣18−6+1∣)/( (√(13))))=((11)/( (√(13)))) The tangent line is 3x+2y+12=0 and 3x+2y−12=0

$${let}\:\mathrm{3}{x}+\mathrm{2}{y}+{c}\:=\:\mathrm{0}\:{is}\:{tangent}\:{to} \\ $$$${hypebola}\:.\:{we}\:{have}\:\frac{\mathrm{1}}{\mathrm{12}}{x}−\frac{\mathrm{1}}{\mathrm{9}}{y}.{y}'=\mathrm{0} \\ $$$$\Rightarrow{y}'=\frac{\frac{\mathrm{3}}{\mathrm{36}}{x}}{\frac{\mathrm{4}}{\mathrm{36}}{y}}\:=\:\frac{\mathrm{3}{x}}{\mathrm{4}{y}}\:=\:−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\Rightarrow{x}=−\mathrm{2}{y}\:{and}\:\frac{\mathrm{4}{y}^{\mathrm{2}} }{\mathrm{24}}−\frac{{y}^{\mathrm{2}} }{\mathrm{18}}\:=\mathrm{1} \\ $$$$\Rightarrow{y}^{\mathrm{2}} =\mathrm{9}\:\Rightarrow\begin{cases}{{y}=\mathrm{3}\:;\:{x}=−\mathrm{6}}\\{{y}=−\mathrm{3}\:;\:{x}=\mathrm{6}}\end{cases} \\ $$$${So}\:{the}\:{contac}\:{point}\:{at}\:\left(−\mathrm{6},\mathrm{3}\right)\:{or}\:\left(\mathrm{6},−\mathrm{3}\right) \\ $$$${for}\:\left(−\mathrm{6},\mathrm{3}\right)\Rightarrow{distance}\:=\frac{\mid−\mathrm{18}+\mathrm{6}+\mathrm{1}\mid}{\:\sqrt{\mathrm{13}}}=\frac{\mathrm{11}}{\:\sqrt{\mathrm{13}}} \\ $$$${for}\:\left(\mathrm{6},−\mathrm{3}\right)\Rightarrow{distance}=\frac{\mid\mathrm{18}−\mathrm{6}+\mathrm{1}\mid}{\:\sqrt{\mathrm{13}}}=\frac{\mathrm{11}}{\:\sqrt{\mathrm{13}}} \\ $$$${The}\:{tangent}\:{line}\:{is}\:\mathrm{3}{x}+\mathrm{2}{y}+\mathrm{12}=\mathrm{0} \\ $$$${and}\:\mathrm{3}{x}+\mathrm{2}{y}−\mathrm{12}=\mathrm{0}\: \\ $$

Commented by horsebrand11 last updated on 12/Feb/23

Commented by Humble last updated on 12/Feb/23

thank you sir

$${thank}\:{you}\:{sir} \\ $$

Answered by mr W last updated on 12/Feb/23

an other method say point P(p,q) on the hyperbola, its distance to line 3x+2y+1=0 is s=((3p+2q+1)/( (√(3^2 +2^2 ))))=((3p+2q+1)/( (√(13)))) (p^2 /(24))−(q^2 /(18))=1 F=((3p+2q+1)/( (√(13))))+λ((p^2 /(24))−(q^2 /(18))−1) (∂F/∂p)=(3/( (√(13))))+((λp)/(12))=0 ⇒p=−((36)/(λ(√(13)))) (∂F/∂q)=(2/( (√(13))))−((λq)/9)=0 ⇒q=((18)/(λ(√(13)))) (1/(24))(−((36)/(λ(√(13)))))^2 −(1/(18))(((18)/(λ(√(13)))))^2 =1 ⇒λ^2 =((36)/(13)) ⇒λ=±(6/( (√(13)))) ⇒p=∓((36)/( (√(13))))×((√(13))/6)=∓6 ⇒q=±((18)/( (√(13))))×((√(13))/6)=±3 ⇒the point is (−6,3) or (6,−3) ⇒s=((∣−3×6+2×3+1∣)/( (√(13))))=((11)/( (√(13))))<(√(13)) ⇒s=((∣3×6−2×3+1∣)/( (√(13))))=((13)/( (√(13))))=(√(13)) point (−6,3) on hyperbola is closest to line 3x+2y+1 with the smallest distance ((11)/( (√(13)))).

$${an}\:{other}\:{method} \\ $$$${say}\:{point}\:{P}\left({p},{q}\right)\:{on}\:{the}\:{hyperbola},\:{its} \\ $$$${distance}\:{to}\:{line}\:\mathrm{3}{x}+\mathrm{2}{y}+\mathrm{1}=\mathrm{0}\:{is} \\ $$$${s}=\frac{\mathrm{3}{p}+\mathrm{2}{q}+\mathrm{1}}{\:\sqrt{\mathrm{3}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} }}=\frac{\mathrm{3}{p}+\mathrm{2}{q}+\mathrm{1}}{\:\sqrt{\mathrm{13}}} \\ $$$$\frac{{p}^{\mathrm{2}} }{\mathrm{24}}−\frac{{q}^{\mathrm{2}} }{\mathrm{18}}=\mathrm{1} \\ $$$${F}=\frac{\mathrm{3}{p}+\mathrm{2}{q}+\mathrm{1}}{\:\sqrt{\mathrm{13}}}+\lambda\left(\frac{{p}^{\mathrm{2}} }{\mathrm{24}}−\frac{{q}^{\mathrm{2}} }{\mathrm{18}}−\mathrm{1}\right) \\ $$$$\frac{\partial{F}}{\partial{p}}=\frac{\mathrm{3}}{\:\sqrt{\mathrm{13}}}+\frac{\lambda{p}}{\mathrm{12}}=\mathrm{0}\:\Rightarrow{p}=−\frac{\mathrm{36}}{\lambda\sqrt{\mathrm{13}}} \\ $$$$\frac{\partial{F}}{\partial{q}}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{13}}}−\frac{\lambda{q}}{\mathrm{9}}=\mathrm{0}\:\Rightarrow{q}=\frac{\mathrm{18}}{\lambda\sqrt{\mathrm{13}}} \\ $$$$\frac{\mathrm{1}}{\mathrm{24}}\left(−\frac{\mathrm{36}}{\lambda\sqrt{\mathrm{13}}}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{18}}\left(\frac{\mathrm{18}}{\lambda\sqrt{\mathrm{13}}}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$$\Rightarrow\lambda^{\mathrm{2}} =\frac{\mathrm{36}}{\mathrm{13}}\:\Rightarrow\lambda=\pm\frac{\mathrm{6}}{\:\sqrt{\mathrm{13}}} \\ $$$$\Rightarrow{p}=\mp\frac{\mathrm{36}}{\:\sqrt{\mathrm{13}}}×\frac{\sqrt{\mathrm{13}}}{\mathrm{6}}=\mp\mathrm{6} \\ $$$$\Rightarrow{q}=\pm\frac{\mathrm{18}}{\:\sqrt{\mathrm{13}}}×\frac{\sqrt{\mathrm{13}}}{\mathrm{6}}=\pm\mathrm{3} \\ $$$$\Rightarrow{the}\:{point}\:{is}\:\left(−\mathrm{6},\mathrm{3}\right)\:{or}\:\left(\mathrm{6},−\mathrm{3}\right) \\ $$$$\Rightarrow{s}=\frac{\mid−\mathrm{3}×\mathrm{6}+\mathrm{2}×\mathrm{3}+\mathrm{1}\mid}{\:\sqrt{\mathrm{13}}}=\frac{\mathrm{11}}{\:\sqrt{\mathrm{13}}}<\sqrt{\mathrm{13}} \\ $$$$\Rightarrow{s}=\frac{\mid\mathrm{3}×\mathrm{6}−\mathrm{2}×\mathrm{3}+\mathrm{1}\mid}{\:\sqrt{\mathrm{13}}}=\frac{\mathrm{13}}{\:\sqrt{\mathrm{13}}}=\sqrt{\mathrm{13}} \\ $$$${point}\:\left(−\mathrm{6},\mathrm{3}\right)\:{on}\:{hyperbola}\:{is} \\ $$$${closest}\:{to}\:{line}\:\mathrm{3}{x}+\mathrm{2}{y}+\mathrm{1}\:{with}\:{the} \\ $$$${smallest}\:{distance}\:\frac{\mathrm{11}}{\:\sqrt{\mathrm{13}}}. \\ $$

Commented by mr W last updated on 12/Feb/23

Commented by Humble last updated on 12/Feb/23

much appreciated sir. and great solution

$${much}\:{appreciated}\:{sir}.\:{and}\:{great}\:{solution} \\ $$

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