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Question Number 112358 by john santu last updated on 07/Sep/20

The point of the curve 3x^2 −4y^2 =72 which nearest to the line 3x+2y=1 is___ (a) (6,3) (c) (6,6) (b) (6,−3) (d) (6,5)

$${The}\:{point}\:{of}\:{the}\:{curve}\:\mathrm{3}{x}^{\mathrm{2}} −\mathrm{4}{y}^{\mathrm{2}} =\mathrm{72} \\ $$$${which}\:{nearest}\:{to}\:{the}\:{line}\:\mathrm{3}{x}+\mathrm{2}{y}=\mathrm{1} \\ $$$${is\_\_\_} \\ $$$$\left({a}\right)\:\left(\mathrm{6},\mathrm{3}\right)\:\:\:\:\:\:\:\left({c}\right)\:\left(\mathrm{6},\mathrm{6}\right) \\ $$$$\left({b}\right)\:\left(\mathrm{6},−\mathrm{3}\right)\:\:\left({d}\right)\:\left(\mathrm{6},\mathrm{5}\right) \\ $$

Answered by MJS_new last updated on 07/Sep/20

distance of point to given line is d=((∣3x+2y−1∣)/( (√(13)))) curve: y=±((√(3x^2 −72))/4) ((∣3x±((√(3x^2 −72))/2)−1∣)/( (√(13)))) the absolute doesn′t change the values only the signs. we need (d/dx)[3x±((√(3x^2 −72))/2)−1]=0 3±((x(√3))/( (√(x^2 −24))))=0 ⇒ x=±6 ⇒ y=±3 inserting in d we get the minimum with x=6∧y=−3

$$\mathrm{distance}\:\mathrm{of}\:\mathrm{point}\:\mathrm{to}\:\mathrm{given}\:\mathrm{line}\:\mathrm{is} \\ $$$${d}=\frac{\mid\mathrm{3}{x}+\mathrm{2}{y}−\mathrm{1}\mid}{\:\sqrt{\mathrm{13}}} \\ $$$$\mathrm{curve}:\:{y}=\pm\frac{\sqrt{\mathrm{3}{x}^{\mathrm{2}} −\mathrm{72}}}{\mathrm{4}} \\ $$$$\frac{\mid\mathrm{3}{x}\pm\frac{\sqrt{\mathrm{3}{x}^{\mathrm{2}} −\mathrm{72}}}{\mathrm{2}}−\mathrm{1}\mid}{\:\sqrt{\mathrm{13}}} \\ $$$$\mathrm{the}\:\mathrm{absolute}\:\mathrm{doesn}'\mathrm{t}\:\mathrm{change}\:\mathrm{the}\:\mathrm{values}\:\mathrm{only} \\ $$$$\mathrm{the}\:\mathrm{signs}.\:\mathrm{we}\:\mathrm{need} \\ $$$$\frac{{d}}{{dx}}\left[\mathrm{3}{x}\pm\frac{\sqrt{\mathrm{3}{x}^{\mathrm{2}} −\mathrm{72}}}{\mathrm{2}}−\mathrm{1}\right]=\mathrm{0} \\ $$$$\mathrm{3}\pm\frac{{x}\sqrt{\mathrm{3}}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{24}}}=\mathrm{0}\:\Rightarrow\:{x}=\pm\mathrm{6}\:\Rightarrow\:{y}=\pm\mathrm{3} \\ $$$$\mathrm{inserting}\:\mathrm{in}\:{d}\:\mathrm{we}\:\mathrm{get}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{with} \\ $$$${x}=\mathrm{6}\wedge{y}=−\mathrm{3} \\ $$

Commented by john santu last updated on 07/Sep/20

thank you prof...

$${thank}\:{you}\:{prof}... \\ $$

Answered by ajfour last updated on 07/Sep/20

tangent to curve ((xx_1 )/(24))−((yy_1 )/(18))=1 slope = ((3x_1 )/(4y_1 ))=−(3/2) ⇒ x_1 =−2y_1 ⇒ 3x_1 ^2 −x_1 ^2 = 72 x_1 = 6 , y_1 =−3

$${tangent}\:{to}\:{curve} \\ $$$$\frac{{xx}_{\mathrm{1}} }{\mathrm{24}}−\frac{{yy}_{\mathrm{1}} }{\mathrm{18}}=\mathrm{1} \\ $$$${slope}\:\:=\:\frac{\mathrm{3}{x}_{\mathrm{1}} }{\mathrm{4}{y}_{\mathrm{1}} }=−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\Rightarrow\:\:{x}_{\mathrm{1}} =−\mathrm{2}{y}_{\mathrm{1}} \\ $$$$\Rightarrow\:\:\mathrm{3}{x}_{\mathrm{1}} ^{\mathrm{2}} −{x}_{\mathrm{1}} ^{\mathrm{2}} =\:\mathrm{72} \\ $$$${x}_{\mathrm{1}} =\:\mathrm{6}\:,\:{y}_{\mathrm{1}} =−\mathrm{3} \\ $$

Commented by ajfour last updated on 07/Sep/20

Commented by bemath last updated on 07/Sep/20

yes...your method same to my solution. by tangent line of hyperbola parallel to given line

$$\mathrm{yes}...\mathrm{your}\:\mathrm{method}\:\mathrm{same}\:\mathrm{to}\:\mathrm{my}\:\mathrm{solution}.\:\mathrm{by}\:\mathrm{tangent} \\ $$$$\mathrm{line}\:\mathrm{of}\:\mathrm{hyperbola}\:\mathrm{parallel}\:\mathrm{to}\:\mathrm{given}\:\mathrm{line} \\ $$

Commented by ajfour last updated on 07/Sep/20

Nice to know; i think you will like attempting Q. 112356 posted by me..

$${Nice}\:{to}\:{know};\:{i}\:{think}\:{you}\:{will}\:{like} \\ $$$${attempting}\:\:{Q}.\:\mathrm{112356}\:{posted}\:{by}\:{me}.. \\ $$

Commented by bemath last updated on 07/Sep/20

yes. your question is greatt

$$\mathrm{yes}.\:\mathrm{your}\:\mathrm{question}\:\mathrm{is}\:\mathrm{greatt} \\ $$

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