Question Number 186952 by Humble last updated on 12/Feb/23
Answered by horsebrand11 last updated on 12/Feb/23
$${let}\:\mathrm{3}{x}+\mathrm{2}{y}+{c}\:=\:\mathrm{0}\:{is}\:{tangent}\:{to} \\ $$$${hypebola}\:.\:{we}\:{have}\:\frac{\mathrm{1}}{\mathrm{12}}{x}−\frac{\mathrm{1}}{\mathrm{9}}{y}.{y}'=\mathrm{0} \\ $$$$\Rightarrow{y}'=\frac{\frac{\mathrm{3}}{\mathrm{36}}{x}}{\frac{\mathrm{4}}{\mathrm{36}}{y}}\:=\:\frac{\mathrm{3}{x}}{\mathrm{4}{y}}\:=\:−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\Rightarrow{x}=−\mathrm{2}{y}\:{and}\:\frac{\mathrm{4}{y}^{\mathrm{2}} }{\mathrm{24}}−\frac{{y}^{\mathrm{2}} }{\mathrm{18}}\:=\mathrm{1} \\ $$$$\Rightarrow{y}^{\mathrm{2}} =\mathrm{9}\:\Rightarrow\begin{cases}{{y}=\mathrm{3}\:;\:{x}=−\mathrm{6}}\\{{y}=−\mathrm{3}\:;\:{x}=\mathrm{6}}\end{cases} \\ $$$${So}\:{the}\:{contac}\:{point}\:{at}\:\left(−\mathrm{6},\mathrm{3}\right)\:{or}\:\left(\mathrm{6},−\mathrm{3}\right) \\ $$$${for}\:\left(−\mathrm{6},\mathrm{3}\right)\Rightarrow{distance}\:=\frac{\mid−\mathrm{18}+\mathrm{6}+\mathrm{1}\mid}{\:\sqrt{\mathrm{13}}}=\frac{\mathrm{11}}{\:\sqrt{\mathrm{13}}} \\ $$$${for}\:\left(\mathrm{6},−\mathrm{3}\right)\Rightarrow{distance}=\frac{\mid\mathrm{18}−\mathrm{6}+\mathrm{1}\mid}{\:\sqrt{\mathrm{13}}}=\frac{\mathrm{11}}{\:\sqrt{\mathrm{13}}} \\ $$$${The}\:{tangent}\:{line}\:{is}\:\mathrm{3}{x}+\mathrm{2}{y}+\mathrm{12}=\mathrm{0} \\ $$$${and}\:\mathrm{3}{x}+\mathrm{2}{y}−\mathrm{12}=\mathrm{0}\: \\ $$
Commented by horsebrand11 last updated on 12/Feb/23
Commented by Humble last updated on 12/Feb/23
$${thank}\:{you}\:{sir} \\ $$
Answered by mr W last updated on 12/Feb/23
$${an}\:{other}\:{method} \\ $$$${say}\:{point}\:{P}\left({p},{q}\right)\:{on}\:{the}\:{hyperbola},\:{its} \\ $$$${distance}\:{to}\:{line}\:\mathrm{3}{x}+\mathrm{2}{y}+\mathrm{1}=\mathrm{0}\:{is} \\ $$$${s}=\frac{\mathrm{3}{p}+\mathrm{2}{q}+\mathrm{1}}{\:\sqrt{\mathrm{3}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} }}=\frac{\mathrm{3}{p}+\mathrm{2}{q}+\mathrm{1}}{\:\sqrt{\mathrm{13}}} \\ $$$$\frac{{p}^{\mathrm{2}} }{\mathrm{24}}−\frac{{q}^{\mathrm{2}} }{\mathrm{18}}=\mathrm{1} \\ $$$${F}=\frac{\mathrm{3}{p}+\mathrm{2}{q}+\mathrm{1}}{\:\sqrt{\mathrm{13}}}+\lambda\left(\frac{{p}^{\mathrm{2}} }{\mathrm{24}}−\frac{{q}^{\mathrm{2}} }{\mathrm{18}}−\mathrm{1}\right) \\ $$$$\frac{\partial{F}}{\partial{p}}=\frac{\mathrm{3}}{\:\sqrt{\mathrm{13}}}+\frac{\lambda{p}}{\mathrm{12}}=\mathrm{0}\:\Rightarrow{p}=−\frac{\mathrm{36}}{\lambda\sqrt{\mathrm{13}}} \\ $$$$\frac{\partial{F}}{\partial{q}}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{13}}}−\frac{\lambda{q}}{\mathrm{9}}=\mathrm{0}\:\Rightarrow{q}=\frac{\mathrm{18}}{\lambda\sqrt{\mathrm{13}}} \\ $$$$\frac{\mathrm{1}}{\mathrm{24}}\left(−\frac{\mathrm{36}}{\lambda\sqrt{\mathrm{13}}}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{18}}\left(\frac{\mathrm{18}}{\lambda\sqrt{\mathrm{13}}}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$$\Rightarrow\lambda^{\mathrm{2}} =\frac{\mathrm{36}}{\mathrm{13}}\:\Rightarrow\lambda=\pm\frac{\mathrm{6}}{\:\sqrt{\mathrm{13}}} \\ $$$$\Rightarrow{p}=\mp\frac{\mathrm{36}}{\:\sqrt{\mathrm{13}}}×\frac{\sqrt{\mathrm{13}}}{\mathrm{6}}=\mp\mathrm{6} \\ $$$$\Rightarrow{q}=\pm\frac{\mathrm{18}}{\:\sqrt{\mathrm{13}}}×\frac{\sqrt{\mathrm{13}}}{\mathrm{6}}=\pm\mathrm{3} \\ $$$$\Rightarrow{the}\:{point}\:{is}\:\left(−\mathrm{6},\mathrm{3}\right)\:{or}\:\left(\mathrm{6},−\mathrm{3}\right) \\ $$$$\Rightarrow{s}=\frac{\mid−\mathrm{3}×\mathrm{6}+\mathrm{2}×\mathrm{3}+\mathrm{1}\mid}{\:\sqrt{\mathrm{13}}}=\frac{\mathrm{11}}{\:\sqrt{\mathrm{13}}}<\sqrt{\mathrm{13}} \\ $$$$\Rightarrow{s}=\frac{\mid\mathrm{3}×\mathrm{6}−\mathrm{2}×\mathrm{3}+\mathrm{1}\mid}{\:\sqrt{\mathrm{13}}}=\frac{\mathrm{13}}{\:\sqrt{\mathrm{13}}}=\sqrt{\mathrm{13}} \\ $$$${point}\:\left(−\mathrm{6},\mathrm{3}\right)\:{on}\:{hyperbola}\:{is} \\ $$$${closest}\:{to}\:{line}\:\mathrm{3}{x}+\mathrm{2}{y}+\mathrm{1}\:{with}\:{the} \\ $$$${smallest}\:{distance}\:\frac{\mathrm{11}}{\:\sqrt{\mathrm{13}}}. \\ $$
Commented by mr W last updated on 12/Feb/23
Commented by Humble last updated on 12/Feb/23
$${much}\:{appreciated}\:{sir}.\:{and}\:{great}\:{solution} \\ $$