Find-the-maximum-distance-between-two-points-on-the-curve-x-4-a-4-y-4-b-4-1-

Question Number 144995 by liberty last updated on 01/Jul/21

Find the maximum distance

between two points on the

curve \frac{x^{4}}{a^{4}} + \frac{y^{4}}{b^{4}} = 1 .

Answered by mr W last updated on 01/Jul/21

d u e t o s y m m e t r y, t h e m a x i m u m

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

c u r v e i s t w o t i m e s o f t h e m a x i m u m

d i s t a n c e f r o m a p o i n t o n t h e c u r v e

t o t h e o r i g i n .

p o i n t P (x, y)

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

x = r \cos θ

y = r \sin θ

\frac{r^{4} \cos^{4} θ}{a^{4}} + \frac{r^{4} \sin^{4} θ}{b^{4}} = 1

\frac{\cos^{4} θ}{a^{2}} + \frac{\sin^{4} θ}{b^{4}} = \frac{1}{r^{4}} = Φ

\frac{d Φ}{d θ} = - \frac{4 \cos^{3} θ \sin θ}{a^{4}} + \frac{4 \sin^{3} θ \cos θ}{b^{4}} = 0

(- \frac{\cos^{2} θ}{a^{4}} + \frac{\sin^{2} θ}{b^{4}}) \cos θ \sin θ = 0

\Rightarrow \sin θ = 0 \Rightarrow θ = 0 \Rightarrow r = a

\Rightarrow \cos θ = 0 \Rightarrow θ = \frac{π}{2} \Rightarrow r = b

\Rightarrow - \frac{\cos^{2} θ}{a^{4}} + \frac{\sin^{2} θ}{b^{4}}

\Rightarrow \tan θ = \frac{b^{2}}{a^{2}}

\Rightarrow \frac{1}{r_{m a x}^{4}} = \frac{1}{a^{4} + b^{4}}

\Rightarrow r_{m a x} = \sqrt[4]{a^{4} + b^{4}} > a, > b

\Rightarrow d_{m a x} = 2 r_{m a x} = 2 \sqrt[4]{a^{4} + b^{4}}

Commented by mr W last updated on 01/Jul/21

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