find-the-point-on-the-curve-3x-2-4y-2-72-closest-to-line-3x-2y-1-0-

Question Number 120378 by john santu last updated on 31/Oct/20

f i n d t h e p o i n t o n t h e c u r v e 3 x^{2} - 4 y^{2} = 72

c l o s e s t t o l i n e 3 x + 2 y - 1 = 0

Commented by bramlexs22 last updated on 31/Oct/20

l e t (x, y) i s a p o i n t o n h y p e r b o l a

y^{2} = \frac{3 x^{2} - 72}{4} a n d c l o s e s t t o l i n e

3 x + 2 y - 1 = 0, w e g e t t h e d i s t a n c e

d = \frac{∣ 3 x \pm 2 \sqrt{\frac{3 x^{2} - 72}{4}} - 1 ∣}{\sqrt{13}}

d = \frac{∣ 3 x \pm \sqrt{3 x^{2} - 72} - 1 ∣}{\sqrt{13}}

w e t a k e \frac{d}{d x} [\frac{3 x \pm \sqrt{3 x^{2} - 72} - 1}{\sqrt{13}}] = 0

\Leftrightarrow 3 \pm \frac{3 x}{\sqrt{3 x^{2} - 72}} = 0

w e g e t x^{2} = 36 o r x = \pm 6 t h e n

y = \pm \sqrt{\frac{3.36 - 72}{4}} = \pm \sqrt{9} = \pm 3

c h e c k p o i n t

(i) (6, 3) \Rightarrow d = \frac{18 + 6 - 1}{\sqrt{13}} = \frac{23}{\sqrt{13}}

(i i) (6, - 3) \Rightarrow d = \frac{18 - 6 - 1}{\sqrt{13}} = \frac{11}{\sqrt{13}} \leftarrow d_{m i n}

(i i i) (- 6, 3) \Rightarrow d = \frac{∣ - 18 + 6 - 1 ∣}{\sqrt{13}} = \frac{17}{\sqrt{13}}

(i v) (- 6, - 3) \Rightarrow d = \frac{∣ - 18 - 6 - 1 ∣}{\sqrt{13}} = \frac{25}{\sqrt{13}}

Commented by john santu last updated on 31/Oct/20

Answered by mr W last updated on 31/Oct/20

3 x^{2} - 4 y^{2} = 72

6 x - 8 y \frac{d y}{d x} = 0

6 x - 8 y \times (- \frac{3}{2}) = 0

\Rightarrow x = - 2 y

3 {(- 2 y)}^{2} - 4 y^{2} = 72

y^{2} = 9

\Rightarrow y = \pm 3

\Rightarrow x = \mp 6

\Rightarrow (6, - 3) a n d (- 6, 3)

\frac{1}{\sqrt{3^{2} + 2^{2}}} ∣ 3 \times 6 + 2 (- 3) - 1 ∣= \frac{11}{\sqrt{13}} ✓

\frac{1}{\sqrt{3^{2} + 2^{2}}} ∣ 3 \times (- 6) + 2 \times 3 - 1 ∣= \frac{13}{\sqrt{13}}

\Rightarrow (6, - 3) i s c l o s e s t t o 3 x + 2 y - 1 = 0 .

a n d t h e c l o s e s t d i s t a n c e i s \frac{11}{\sqrt{13}} .

Commented by mr W last updated on 31/Oct/20

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