Question-50820

Question Number 50820 by ajfour last updated on 20/Dec/18

Commented by ajfour last updated on 20/Dec/18

C h o o s e y o u r o r i g i n a n d f i n d

e q u a t i o n o f p a r a b o l a s u c h t h a t

i t h a s m a x i m u m l e n g t h i n s i d e

e l l i p s e (p a r a m e t e r s a a n d b) .

Answered by ajfour last updated on 21/Dec/18

y = {A x}^{2} - b

b \sin θ = {A a}^{2} \cos^{2} θ - b

l e t \sin θ = t

\Rightarrow {A a}^{2} (1 - t^{2}) - b = b t

o t h e r t h a n t = - 1

1 - t = \frac{b}{{A a}^{2}} \Rightarrow t = 1 - \frac{b}{{A a}^{2}}

\Rightarrow b \sin θ = y_{A} = b - \frac{b^{2}}{{A a}^{2}}

x = \sqrt{\frac{b + y}{A}} \Rightarrow \frac{d x}{d y} = \frac{1}{2 \sqrt{A (b + y)}}

L = \int_{A}^{C} d l = 2 \int_{- b}^{y_{A}} \sqrt{1 + {(\frac{d x}{d y})}^{2}} d y

= 2 \int_{- b}^{y_{A}} \sqrt{1 + \frac{1}{4 A (b + y)}} d y

\frac{1}{2} \frac{d L}{d A} = \int_{- b}^{y_{A}} \frac{(- \frac{1}{4 A^{2} (b + y)})}{2 \sqrt{1 + \frac{1}{4 A (b + y)}}} d y

+ \frac{b^{2}}{A^{2} a^{2}} \sqrt{1 + \frac{1}{4 A (2 b - \frac{b^{2}}{{A a}^{2}})}}

l e t I = \int_{- b}^{y_{A}} \frac{\frac{1}{(b + y)}}{\sqrt{1 + \frac{1}{4 A (b + y)}}} d y

= \int \frac{d z}{\sqrt{z^{2} + \frac{z}{4 A}}} = \int \frac{d z}{\sqrt{{(z + \frac{1}{2 A})}^{2} - \frac{1}{4 A^{2}}}}

= \ln ∣ z + c + \sqrt{{(z + c)}^{2} - c^{2}} ∣ + k

w h e r e [c = \frac{1}{2 A}]

y_{A} = b - \frac{b^{2}}{{A a}^{2}}; z_{A} = b + y_{A} = 2 b - \frac{2 b^{2} c}{a^{2}}

y_{B} = - b \Rightarrow z_{B} = 0

l e t z_{A} + c = t_{A} = 2 b + \frac{1}{2 A} (1 - \frac{2 b^{2}}{a^{2}})

t_{B} = z_{B} + c = \frac{1}{2 A}

I = \ln \frac{t_{A} + \sqrt{t_{A}^{2} - c^{2}}}{t_{B}}

\frac{d L}{d A} = 0 \Rightarrow

\frac{I}{8 A^{2}} = \frac{b^{2}}{A^{2} a^{2}} \sqrt{1 + \frac{1}{4 A (2 b - \frac{b^{2}}{{A a}^{2}})}}

\Rightarrow \ln \frac{t_{A} + \sqrt{t_{A}^{2} - c^{2}}}{t_{B}} = \frac{8 b^{2}}{a^{2}} \sqrt{1 + \frac{1}{4 {A t}_{A}}}

\Rightarrow I n h e r e c = \frac{1}{2 A}

w i t h t_{A} = 2 b + \frac{1}{2 A} (1 - \frac{2 b^{2}}{a^{2}}); t_{B} = \frac{1}{2 A}

A i s o b t a i n a b l e f r o m a b o v e e q u a t i o n .

I f f o r e x a m p l e A = 0

\Rightarrow \ln (1 - \frac{2 b^{2}}{a^{2}} + \sqrt{{(1 - \frac{2 b^{2}}{a^{2}})}^{2} - 1})

= \frac{8 b^{2}}{a^{2}} \sqrt{1 + 2 (1 - \frac{2 b^{2}}{a^{2}})}

\Rightarrow b = 0 (n a t u r a l l y i f w i t h i n a n

e l l i p s e w i t h b \to 0, p a r a b o l a s h a l l

h a v e m a x i m u m l e n g t h i f A = 0 .

Commented by ajfour last updated on 21/Dec/18

m r W S i r c a n y o u c a s t a g l a n c e

p l e a s e . .

Commented by mr W last updated on 21/Dec/18

p l e a s e c h e c k w h a t i s c i n

\Rightarrow \ln \frac{t_{A} + \sqrt{t_{A}^{2} - c^{2}}}{t_{B}} = \frac{8 b^{2}}{a^{2}} \sqrt{1 + \frac{1}{4 {A t}_{A}}}

i n t h i s e q n . w e s h o u l d o n l y h a v e a, b, A .

Commented by ajfour last updated on 21/Dec/18

T h a n k y o u S i r, c = \frac{1}{2 A} .

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