find-the-point-on-the-graph-of-f-x-1-x-2-that-are-closest-to-O-0-0-

Question Number 131188 by abdurehime last updated on 02/Feb/21

find the point on the graph of f (x) = 1 - x^{2}

that are closest to O (0, 0)

Answered by john_santu last updated on 02/Feb/21

l e t P (x, y) i s t h e p o i n t o n t h e

c u r v e .

l e t t h e d i s t a n c e b e t w e e n p o i n t P t o p o i n t O

= d = \sqrt{x^{2} + y^{2}} = \sqrt{1 - y + y^{2}}

o r d^{2} = y^{2} - y + 1; y^{2} - y + 1 - d^{2} = 0

b y t a n g e n c y Δ = 0

\Rightarrow 1 - 4 (1 - d^{2}) = 0; 1 - d^{2} = \frac{1}{4}

\Rightarrow d^{2} = \frac{3}{4} \Rightarrow y^{2} - y + \frac{1}{4} = 0

{(y - \frac{1}{2})}^{2} = 0 \to {\begin{cases} y = \frac{1}{2} \\ x = \pm \frac{1}{2} \sqrt{2} \end{cases}

w e g e t \to {\begin{cases} P_{1} (\frac{1}{2} \sqrt{2}, \frac{1}{2}) \\ P_{2} (- \frac{1}{2} \sqrt{2}, \frac{1}{2}) \end{cases}

Commented by abdurehime last updated on 02/Feb/21

i satisfied 100 % allh bless you

Answered by mr W last updated on 02/Feb/21

s a y t h e p o i n t i s (x, y) w i t h

y = 1 - x^{2}

t h e d i s t a n c e f r o m (x, y) t o (0, 0) i s d .

d^{2} = x^{2} + y^{2} = x^{2} + {(1 - x^{2})}^{2}

= x^{4} - x^{2} + 1

= {(x^{2} - \frac{1}{2})}^{2} + \frac{3}{4} ⩾ \frac{3}{4}

\Rightarrow d_{m i n} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}

a t x^{2} = \frac{1}{2} \Rightarrow x = \pm \frac{\sqrt{2}}{2}, y = 1 - \frac{1}{2} = \frac{1}{2}

\Rightarrow p o i n t (- \frac{\sqrt{2}}{2}, \frac{1}{2}) o r p o i n t (\frac{\sqrt{2}}{2}, \frac{1}{2})

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