The-nearest-distance-of-0-3-to-curve-y-6-x-2-is-

Question Number 83445 by naka3546 last updated on 02/Mar/20

T h e n e a r e s t d i s t a n c e o f (0, 3) t o c u r v e : y = 6 - x^{2} i s \dots

Commented by MJS last updated on 02/Mar/20

P = (\begin{matrix} x \\ y \end{matrix}) = (\begin{matrix} x \\ 6 - x^{2} \end{matrix}); {distance}^{2} is x^{2} + {(3 - x^{2})}^{2} = x^{4} - 5 x^{2} + 9; \frac{d}{d x} [x^{4} - 5 x^{2} + 9] = 0 \Leftrightarrow x (x^{2} - \frac{5}{2}) = 0 \Rightarrow x = 0 \lor x = \pm \frac{\sqrt{10}}{2} \Rightarrow \min at x = \pm \frac{\sqrt{10}}{2} \Rightarrow \min dist . = \frac{\sqrt{11}}{2}

Answered by john santu last updated on 02/Mar/20

d = \sqrt{x^{2} + {(3 - x^{2})}^{2}}

let f (x) = x^{2} + {(3 - x^{2})}^{2}

f^{'} (x) = 2 x - 2 . 2 x (3 - x^{2}) = 0

2 x (1 - 2 (3 - x^{2})) = 0

\Rightarrow x = 0 \land x^{2} = \frac{5}{2}

for x = 0 \Rightarrow d = \sqrt{9} = 3 = \sqrt{\frac{12}{4}}

for x^{2} = \frac{5}{2} \Rightarrow d = \sqrt{\frac{5}{2} + \frac{1}{4}} = \sqrt{\frac{11}{4}}

d_{\min} = \frac{1}{2} \sqrt{11} \Leftarrow the answer

Answered by mr W last updated on 02/Mar/20

y = 6 - x^{2}

x^{2} + {(y - 3)}^{2} = d^{2}

y^{2} - 7 y + 15 - d^{2} = 0

t a n g e n c y :

Δ = 7^{2} - 4 (15 - d^{2}) = 0

\Rightarrow d = \sqrt{15 - \frac{49}{4}} = \frac{\sqrt{11}}{2}

Commented by john santu last updated on 03/Mar/20

what is tangency sir ?

Commented by mr W last updated on 03/Mar/20

i f t h e c i r c l e x^{2} + {(y - 3)}^{2} = d^{2} t a n g e n t s

t h e p a r a b o l a y = 6 - x^{2}, t h e n r a d i u s d

o f t h e c i r c l e i s t h e s m a l l e s t d i s t a n c e

f r o m t h e p o i n t (0, 3) t o c u r v e y = 6 - x^{2} .

w h e n b o t h c u r v e s t a n g e n t e a c h o t h e r,

t h e e q u a t i o n y^{2} - 7 y + 15 - d^{2} = 0 h a s

o n e a n d o n l y o n e s o l u t i o n, t h i s i s

w h a t “ {t a n g e n c y}^{″} m e a n s .

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