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)


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

$T h e p o i n t o f t h e c u r v e 3 x^{2} - 4 y^{2} = 72$ $w h i c h n e a r e s t t o t h e l i n e 3 x + 2 y = 1$ $i s___$ $(a) (6, 3) (c) (6, 6)$ $(b) (6, - 3) (d) (6, 5)$
	Answered by MJS_new last updated on 07/Sep/20

	$distance of point to given line is$ $d = \frac{∣ 3 x + 2 y - 1 ∣}{\sqrt{13}}$ $curve : y = \pm \frac{\sqrt{3 x^{2} - 72}}{4}$ $\frac{∣ 3 x \pm \frac{\sqrt{3 x^{2} - 72}}{2} - 1 ∣}{\sqrt{13}}$ $the absolute {doesn}^{'} t change the values only$ $the signs . we need$ $\frac{d}{d x} [3 x \pm \frac{\sqrt{3 x^{2} - 72}}{2} - 1] = 0$ $3 \pm \frac{x \sqrt{3}}{\sqrt{x^{2} - 24}} = 0 \Rightarrow x = \pm 6 \Rightarrow y = \pm 3$ $inserting in d we get the minimum with$ $x = 6 \land y = - 3$
	Commented by john santu last updated on 07/Sep/20

	$t h a n k y o u p r o f . . .$
	Answered by ajfour last updated on 07/Sep/20

	$t a n g e n t t o c u r v e$ $\frac{{x x}_{1}}{24} - \frac{{y y}_{1}}{18} = 1$ $s l o p e = \frac{3 x_{1}}{4 y_{1}} = - \frac{3}{2}$ $\Rightarrow x_{1} = - 2 y_{1}$ $\Rightarrow 3 x_{1}^{2} - x_{1}^{2} = 72$ $x_{1} = 6, y_{1} = - 3$
	Commented by ajfour last updated on 07/Sep/20

	Commented by bemath last updated on 07/Sep/20

	$yes . . . your method same to my solution . by tangent$ $line of hyperbola parallel to given line$
	Commented by ajfour last updated on 07/Sep/20

	$N i c e t o k n o w; i t h i n k y o u w i l l l i k e$ $a t t e m p t i n g Q . 112356 p o s t e d b y m e . .$
	Commented by bemath last updated on 07/Sep/20

	$yes . your question is greatt$
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