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

Answered by horsebrand11 last updated on 12/Feb/23

$$let3x+2y+c=0istangentto$$$$hypebola.wehave\frac{1}{12}x-\frac{1}{9}y.y^{\prime }=0$$$$\Rightarrow y^{\prime }=\frac{\frac{3}{36}x}{\frac{4}{36}y}=\frac{3x}{4y}=-\frac{3}{2}$$$$\Rightarrow x=-2yand\frac{4y^2}{24}-\frac{y^2}{18}=1$$$$\Rightarrow y^2=9\Rightarrow \{\begin{array}{l}y=3;x=-6\\ y=-3;x=6\end{array}$$$$Sothecontacpointat(-6,3)or(6,-3)$$$$for(-6,3)\Rightarrow distance=\frac{\mid -18+6+1\mid }{\sqrt{13}}=\frac{11}{\sqrt{13}}$$$$for(6,-3)\Rightarrow distance=\frac{\mid 18-6+1\mid }{\sqrt{13}}=\frac{11}{\sqrt{13}}$$$$Thetangentlineis3x+2y+12=0$$$$and3x+2y-12=0$$

Commented by horsebrand11 last updated on 12/Feb/23

Commented by Humble last updated on 12/Feb/23

$$thankyousir$$

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

$$anothermethod$$$$saypointP(p,q)onthehyperbola,its$$$$distancetoline3x+2y+1=0is$$$$s=\frac{3p+2q+1}{\sqrt{3^2+2^2}}=\frac{3p+2q+1}{\sqrt{13}}$$$$\frac{p^2}{24}-\frac{q^2}{18}=1$$$$F=\frac{3p+2q+1}{\sqrt{13}}+\lambda (\frac{p^2}{24}-\frac{q^2}{18}-1)$$$$\frac{\partial F}{\partial p}=\frac{3}{\sqrt{13}}+\frac{\lambda p}{12}=0\Rightarrow p=-\frac{36}{\lambda \sqrt{13}}$$$$\frac{\partial F}{\partial q}=\frac{2}{\sqrt{13}}-\frac{\lambda q}{9}=0\Rightarrow q=\frac{18}{\lambda \sqrt{13}}$$$$\frac{1}{24}{(-\frac{36}{\lambda \sqrt{13}})}^2-\frac{1}{18}{\left(\frac{18}{\lambda \sqrt{13}}\right)}^2=1$$$$\Rightarrow {\lambda }^2=\frac{36}{13}\Rightarrow \lambda =\pm \frac{6}{\sqrt{13}}$$$$\Rightarrow p=\mp \frac{36}{\sqrt{13}}\times \frac{\sqrt{13}}{6}=\mp 6$$$$\Rightarrow q=\pm \frac{18}{\sqrt{13}}\times \frac{\sqrt{13}}{6}=\pm 3$$$$\Rightarrow thepointis(-6,3)or(6,-3)$$$$\Rightarrow s=\frac{\mid -3\times 6+2\times 3+1\mid }{\sqrt{13}}=\frac{11}{\sqrt{13}}<\sqrt{13}$$$$\Rightarrow s=\frac{\mid 3\times 6-2\times 3+1\mid }{\sqrt{13}}=\frac{13}{\sqrt{13}}=\sqrt{13}$$$$point(-6,3)onhyperbolais$$$$closesttoline3x+2y+1withthe$$$$smallestdistance\frac{11}{\sqrt{13}}.$$

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

Commented by Humble last updated on 12/Feb/23

$$muchappreciatedsir.andgreatsolution$$

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