Question-50898

Question Number 50898 by ajfour last updated on 21/Dec/18

Commented by ajfour last updated on 21/Dec/18

I f l e n g t h o f B P i s m a x i m u m

a n d e q u a l t o l, a n d t h e t w o c o l o u r e d

a r e a s e q u a l, f i n d a a n d b o f e l l i p s e .

Answered by ajfour last updated on 21/Dec/18

i g o t a = l \sqrt{2 - \sqrt{2}}, b = l \sqrt{3 \sqrt{2} - 4} .

Commented by mr W last updated on 22/Dec/18

i g o t t h e s a m e s i r .

Answered by tanmay.chaudhury50@gmail.com last updated on 22/Dec/18

a r e a s k y = \frac{1}{2} \times a c o s θ \times b s i n θ + \int_{b s i n θ}^{b} \frac{a}{b} \sqrt{b^{2} - y^{2}} d y

= \frac{a}{b} [∣ \frac{y}{2} \sqrt{b^{2} - y^{2}} + \frac{b^{2}}{2} {s i n}^{- 1} (\frac{y}{b}) ∣_{b s i n θ}^{b}] + \frac{1}{4} s i n 2 θ

= \frac{a}{b} [(\frac{b^{2}}{2} \times \frac{π}{2}) - (\frac{b s i n θ \times b c o s θ}{2} + \frac{b^{2}}{2} θ)] + \frac{s i n 2 θ}{4}

= \frac{π a b}{4} - \frac{a b s i n 2 θ}{4} + \frac{a b θ}{2} + \frac{s i n 2 θ}{4}

g i v e n A_{s} = A_{p} = \frac{π a b}{8} a t θ = θ_{0}

\frac{π a b}{8} = \frac{π a b}{4} - \frac{a b s i n 2 θ_{0}}{4} + \frac{a b θ_{0}}{2} + \frac{s i n 2 θ_{0}}{4}

s i n 2 θ_{0} \times \frac{1}{4} (a b - 1) - \frac{a b}{2} {s i n}^{- 1} (\frac{b^{2}}{a^{2} - b^{2}}) = \frac{π a b}{8}

l e n g t h B P = \sqrt{{(a c o s θ - 0)}^{2} + {(b s i n θ + b)}^{2}} = s

s_{m a x} = l g i v e n

s^{2} = a^{2} {c o s}^{2} θ + b^{2} {(1 + s i n θ)}^{2}

2 s \times \frac{d s}{d θ} = a^{2} \times - s i n 2 θ + b^{2} \times 2 (1 + s i n θ) (c o s θ)

2 b^{2} c o s θ + b^{2} s i n 2 θ - a^{2} s i n 2 θ = 0

c o s θ [2 b^{2} + 2 b^{2} s i n θ - 2 a^{2} s i n θ] = 0

c o s θ = 0 θ = \frac{π}{2} \leftarrow i g n o r e d

s i n θ_{0} = \frac{b^{2}}{a^{2} - b^{2}} c o s θ_{0} = \frac{\sqrt{{(a^{2} - b^{2})}^{2} - b^{4}}}{(a^{2} - b^{2})} = \frac{\sqrt{a^{4} - 2 a^{2} b^{2}}}{(a^{2} - b^{2})}

w a i t \dots

Answered by mr W last updated on 22/Dec/18

Commented by OTCHRRE ABDULLAI last updated on 22/Dec/18

M y m a n y o u a r e t o o g o o d

p l e a s e y o u a r e f r o m w h i c h c o u n t r y ?

Commented by mr W last updated on 22/Dec/18

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

A_{r e d} = A_{b l u e} = \frac{π a b}{8}

A_{r e d} = \int_{0}^{θ} \frac{r^{2} d θ}{2} = \frac{a^{2} b^{2}}{2} \int_{0}^{θ} \frac{d θ}{a^{2} \sin^{2} θ + b^{2} \cos^{2} θ}

= \frac{a^{2} b^{2}}{2} \times \frac{\tan^{- 1} (\frac{a}{b} \tan θ)}{a b}

= \frac{a b}{2} \times \tan^{- 1} (\frac{a}{b} \tan θ) = \frac{π a b}{8}

\Rightarrow \tan^{- 1} (\frac{a}{b} \tan θ) = \frac{π}{4}

\Rightarrow \frac{a}{b} \tan θ = 1

\Rightarrow \tan θ = \frac{b}{a} \Rightarrow \sin θ = \frac{b}{\sqrt{a^{2} + b^{2}}} a n d \cos θ = \frac{a}{\sqrt{a^{2} + b^{2}}}

P (r, θ)

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

\frac{x}{a^{2}} + \frac{y}{b^{2}} y^{'} = 0

\frac{r \cos θ}{a^{2}} + \frac{r \sin θ}{b^{2}} y^{'} = 0

\Rightarrow y^{'} (a t P) = - \frac{b^{2}}{a^{2} \tan θ} = - \frac{b}{a} = - \tan α = - \tan θ

\Rightarrow α = θ

φ = \frac{π}{2} - α = \frac{π}{2} - θ

r \cos θ = l \cos φ = l \sin θ

\Rightarrow r = l \tan θ = \frac{l b}{a}

r \sin θ = l \sin φ - b = l \cos θ - b

\Rightarrow r = \frac{l}{\tan θ} - \frac{b}{\sin θ}

\Rightarrow \frac{l}{\tan θ} - \frac{b}{\sin θ} = l \tan θ

\Rightarrow b = l (\frac{1}{\tan θ} - \tan θ) \sin θ = l \times \frac{a^{2} - b^{2}}{a b} \times \frac{b}{\sqrt{a^{2} + b^{2}}}

\Rightarrow l = \frac{a b \sqrt{a^{2} + b^{2}}}{a^{2} - b^{2}}

r^{2} = \frac{l^{2} b^{2}}{a^{2}} = \frac{b^{2}}{a^{2}} \times \frac{a^{2} b^{2} (a^{2} + b^{2})}{{(a^{2} - b^{2})}^{2}} = \frac{b^{4} (a^{2} + b^{2})}{{(a^{2} - b^{2})}^{2}} = \frac{a^{2} + b^{2}}{2}

\Rightarrow \frac{b^{4}}{{(a^{2} - b^{2})}^{2}} = \frac{1}{2}

{(\sqrt{2} b^{2})}^{2} - {(a^{2} - b^{2})}^{2} = 0

[(\sqrt{2} + 1) b^{2} - a^{2}] [(\sqrt{2} - 1) b^{2} + a^{2}] = 0

\Rightarrow a^{2} = (\sqrt{2} + 1) b^{2}

\Rightarrow l = \frac{a b \sqrt{a^{2} + b^{2}}}{a^{2} - b^{2}} = \frac{a b \sqrt{(2 + \sqrt{2}) b^{2}}}{\sqrt{2} b^{2}} = a \sqrt{\frac{2 + \sqrt{2}}{2}}

\Rightarrow a = \sqrt{\frac{2}{2 + \sqrt{2}}} l = \sqrt{2 - \sqrt{2}} l

\Rightarrow b = \frac{a}{\sqrt{\sqrt{2} + 1}} = \sqrt{\frac{2 - \sqrt{2}}{\sqrt{2} + 1}} l = \sqrt{3 \sqrt{2} - 4} l

Commented by ajfour last updated on 22/Dec/18

W O W!

Commented by mr W last updated on 22/Dec/18

Commented by mr W last updated on 22/Dec/18

d i a g r a m s h o w s t h e c a s e l = 5 .

Commented by ajfour last updated on 22/Dec/18

T o o G o o d S i r, T h a n k s f o r c o n f i r m i n g .

Commented by OTCHRRE ABDULLAI last updated on 23/Dec/18

Y o u a r e w o r l d b e s t m y m a n

i f m a t h e m a t i c s i s f o o t b a l l y o u w i l l

b e m e s s i o r R o n a l d o i r e a l l y l i k e

y o u i n f a c t y o u a r e n o w m y b e s t

f r i e n d

Commented by peter frank last updated on 22/Dec/18

a n d p h y s i c s t o o