Question-208023

Question Number 208023 by efronzo1 last updated on 02/Jun/24

Answered by mr W last updated on 02/Jun/24

x = a + ∣ r ∣ \cos θ

\Rightarrow y = b + ∣ r ∣ \sin θ

\sqrt{x^{2} + y^{2}}

= \sqrt{a^{2} + b^{2} + r^{2} + 2 ∣ r ∣ (a \cos θ + b \sin θ)}

= \sqrt{a^{2} + b^{2} + r^{2} + 2 ∣ r ∣ \sqrt{a^{2} + b^{2}} \cos (θ - \tan^{- 1} \frac{b}{a})}

\Rightarrow {(\sqrt{x^{2} + y^{2}})}_{m a x} = \sqrt{a^{2} + b^{2} + r^{2} + 2 ∣ r ∣ \sqrt{a^{2} + b^{2}}}

= \sqrt{a^{2} + b^{2}} + ∣ r ∣

\Rightarrow {(\sqrt{x^{2} + y^{2}})}_{m i n} = \sqrt{a^{2} + b^{2} + r^{2} - 2 ∣ r ∣ \sqrt{a^{2} + b^{2}}}

=∣ \sqrt{a^{2} + b^{2}} - ∣ r ∣∣

Commented by efronzo1 last updated on 02/Jun/24

when \underset{―}{⋖} \neq 0, if ⋖

Commented by mr W last updated on 02/Jun/24

{i t}^{'} s v a l i d f o r a n y a, b \in R .

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