Question Number 120378 by john santu last updated on 31/Oct/20
$${find}\:{the}\:{point}\:{on}\:{the}\:{curve}\:\mathrm{3}{x}^{\mathrm{2}} −\mathrm{4}{y}^{\mathrm{2}} =\mathrm{72} \\ $$$${closest}\:{to}\:{line}\:\mathrm{3}{x}+\mathrm{2}{y}−\mathrm{1}=\mathrm{0} \\ $$
Commented by bramlexs22 last updated on 31/Oct/20
![let (x,y) is a point on hyperbola y^2 =((3x^2 −72)/4) and closest to line 3x+2y−1=0 , we get the distance d =((∣3x±2(√((3x^2 −72)/4))−1∣)/( (√(13)))) d = ((∣3x±(√(3x^2 −72))−1∣)/( (√(13)))) we take (d/dx) [ ((3x±(√(3x^2 −72))−1)/( (√(13)))) ]=0 ⇔ 3 ± ((3x)/( (√(3x^2 −72)))) = 0 we get x^2 =36 or x=±6 then y=±(√((3.36−72)/4))=±(√9) = ± 3 check point (i) (6,3)⇒d = ((18+6−1)/( (√(13))))=((23)/( (√(13)))) (ii)(6,−3)⇒d=((18−6−1)/( (√(13))))=((11)/( (√(13))))←d_(min) (iii)(−6,3)⇒d=((∣−18+6−1∣)/( (√(13))))=((17)/( (√(13)))) (iv)(−6,−3)⇒d=((∣−18−6−1∣)/( (√(13))))=((25)/( (√(13))))](Q120381.png)
$${let}\:\left({x},{y}\right)\:{is}\:{a}\:{point}\:{on}\:{hyperbola} \\ $$$${y}^{\mathrm{2}} =\frac{\mathrm{3}{x}^{\mathrm{2}} −\mathrm{72}}{\mathrm{4}}\:{and}\:{closest}\:{to}\:{line} \\ $$$$\mathrm{3}{x}+\mathrm{2}{y}−\mathrm{1}=\mathrm{0}\:,\:{we}\:{get}\:{the}\:{distance} \\ $$$${d}\:=\frac{\mid\mathrm{3}{x}\pm\mathrm{2}\sqrt{\frac{\mathrm{3}{x}^{\mathrm{2}} −\mathrm{72}}{\mathrm{4}}}−\mathrm{1}\mid}{\:\sqrt{\mathrm{13}}} \\ $$$${d}\:=\:\frac{\mid\mathrm{3}{x}\pm\sqrt{\mathrm{3}{x}^{\mathrm{2}} −\mathrm{72}}−\mathrm{1}\mid}{\:\sqrt{\mathrm{13}}} \\ $$$${we}\:{take}\:\frac{{d}}{{dx}}\:\left[\:\frac{\mathrm{3}{x}\pm\sqrt{\mathrm{3}{x}^{\mathrm{2}} −\mathrm{72}}−\mathrm{1}}{\:\sqrt{\mathrm{13}}}\:\right]=\mathrm{0} \\ $$$$\Leftrightarrow\:\mathrm{3}\:\pm\:\frac{\mathrm{3}{x}}{\:\sqrt{\mathrm{3}{x}^{\mathrm{2}} −\mathrm{72}}}\:=\:\mathrm{0} \\ $$$${we}\:{get}\:{x}^{\mathrm{2}} =\mathrm{36}\:{or}\:{x}=\pm\mathrm{6}\:{then} \\ $$$${y}=\pm\sqrt{\frac{\mathrm{3}.\mathrm{36}−\mathrm{72}}{\mathrm{4}}}=\pm\sqrt{\mathrm{9}}\:=\:\pm\:\mathrm{3} \\ $$$${check}\:{point} \\ $$$$\left({i}\right)\:\left(\mathrm{6},\mathrm{3}\right)\Rightarrow{d}\:=\:\frac{\mathrm{18}+\mathrm{6}−\mathrm{1}}{\:\sqrt{\mathrm{13}}}=\frac{\mathrm{23}}{\:\sqrt{\mathrm{13}}} \\ $$$$\left({ii}\right)\left(\mathrm{6},−\mathrm{3}\right)\Rightarrow{d}=\frac{\mathrm{18}−\mathrm{6}−\mathrm{1}}{\:\sqrt{\mathrm{13}}}=\frac{\mathrm{11}}{\:\sqrt{\mathrm{13}}}\leftarrow{d}_{{min}} \\ $$$$\left({iii}\right)\left(−\mathrm{6},\mathrm{3}\right)\Rightarrow{d}=\frac{\mid−\mathrm{18}+\mathrm{6}−\mathrm{1}\mid}{\:\sqrt{\mathrm{13}}}=\frac{\mathrm{17}}{\:\sqrt{\mathrm{13}}} \\ $$$$\left({iv}\right)\left(−\mathrm{6},−\mathrm{3}\right)\Rightarrow{d}=\frac{\mid−\mathrm{18}−\mathrm{6}−\mathrm{1}\mid}{\:\sqrt{\mathrm{13}}}=\frac{\mathrm{25}}{\:\sqrt{\mathrm{13}}} \\ $$
Commented by john santu last updated on 31/Oct/20
Answered by mr W last updated on 31/Oct/20
$$\mathrm{3}{x}^{\mathrm{2}} −\mathrm{4}{y}^{\mathrm{2}} =\mathrm{72} \\ $$$$\mathrm{6}{x}−\mathrm{8}{y}\frac{{dy}}{{dx}}=\mathrm{0} \\ $$$$\mathrm{6}{x}−\mathrm{8}{y}×\left(−\frac{\mathrm{3}}{\mathrm{2}}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}=−\mathrm{2}{y} \\ $$$$\mathrm{3}\left(−\mathrm{2}{y}\right)^{\mathrm{2}} −\mathrm{4}{y}^{\mathrm{2}} =\mathrm{72} \\ $$$${y}^{\mathrm{2}} =\mathrm{9} \\ $$$$\Rightarrow{y}=\pm\mathrm{3} \\ $$$$\Rightarrow{x}=\mp\mathrm{6} \\ $$$$\Rightarrow\left(\mathrm{6},−\mathrm{3}\right)\:{and}\:\left(−\mathrm{6},\mathrm{3}\right) \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} }}\mid\mathrm{3}×\mathrm{6}+\mathrm{2}\left(−\mathrm{3}\right)−\mathrm{1}\mid=\frac{\mathrm{11}}{\:\sqrt{\mathrm{13}}}\checkmark \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} }}\mid\mathrm{3}×\left(−\mathrm{6}\right)+\mathrm{2}×\mathrm{3}−\mathrm{1}\mid=\frac{\mathrm{13}}{\:\sqrt{\mathrm{13}}} \\ $$$$\Rightarrow\left(\mathrm{6},−\mathrm{3}\right)\:{is}\:{closest}\:{to}\:\mathrm{3}{x}+\mathrm{2}{y}−\mathrm{1}=\mathrm{0}. \\ $$$${and}\:{the}\:{closest}\:{distance}\:{is}\:\frac{\mathrm{11}}{\:\sqrt{\mathrm{13}}}. \\ $$
Commented by mr W last updated on 31/Oct/20
