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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 \dots

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 \dots 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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