Question-131420 – Tinku Tara

Question Number 131420 by ajfour last updated on 04/Feb/21

Commented by ajfour last updated on 04/Feb/21

I n t e r m s o f e l l i p s e p a r a m e t e r s

a a n d b, f i n d r a d i u s o f t h e

e q u a l r a d i i c i r c l e s .

Answered by mr W last updated on 04/Feb/21

Commented by mr W last updated on 05/Feb/21

l e t μ = \frac{b}{a}

s a y P (a + a \cos ϕ, b + b \sin ϕ)

\tan θ = \frac{b}{a \tan ϕ} = \frac{μ}{\tan ϕ}

\tan ϕ = \frac{μ}{\tan θ}

\sin ϕ = \frac{μ}{\sqrt{μ^{2} + \tan^{2} θ}}

\cos ϕ = \frac{\tan θ}{\sqrt{μ^{2} + \tan^{2} θ}}

y = b + b \sin ϕ - \frac{μ}{\tan ϕ} (x - a - a \cos ϕ)

0 = b (1 + \sin ϕ) - \frac{μ}{\tan ϕ} (x_{A} - a (1 + \cos ϕ))

(1 + \sin ϕ) = \frac{1}{\tan ϕ} (\frac{x_{A}}{a} - (1 + \cos ϕ))

\Rightarrow \frac{x_{A}}{a} = 1 + \cos ϕ + (1 + \sin ϕ) \tan ϕ

\Rightarrow \frac{x_{A}}{a} = 1 + \frac{\tan θ}{\sqrt{μ^{2} + \tan^{2} θ}} + (1 + \frac{μ}{\sqrt{μ^{2} + \tan^{2} θ}}) \frac{μ}{\tan θ}

\Rightarrow \frac{x_{A}}{a} = 1 + \frac{μ}{\tan θ} + \sqrt{1 + {(\frac{μ}{\tan θ})}^{2}}

s a y Q (a + a \cos α, b - b \sin α)

\tan φ = \frac{μ}{\tan α}

\tan α = \frac{μ}{\tan φ}

\sin α = \frac{μ}{\sqrt{μ^{2} + \tan^{2} φ}}

\cos α = \frac{\tan φ}{\sqrt{μ^{2} + \tan^{2} φ}}

x_{C} = a + a \cos α + r \sin φ = x_{A} - \frac{r}{\tan \frac{θ}{2}}

1 + \cos α + λ \sin φ = 1 + \frac{μ}{\tan θ} + \sqrt{1 + {(\frac{μ}{\tan θ})}^{2}} - \frac{λ}{\tan \frac{θ}{2}}

l e t λ = \frac{r}{a}

\frac{\tan φ}{\sqrt{μ^{2} + \tan^{2} φ}} + λ (\sin φ + \frac{1}{\tan \frac{θ}{2}}) = \frac{μ}{\tan θ} + \sqrt{1 + {(\frac{μ}{\tan θ})}^{2}}

y_{C} = b - b \sin α - r \cos φ = r

μ (1 - \sin α) = λ (1 + \cos φ)

λ = \frac{r}{a} = \frac{μ}{1 + \cos φ} (1 - \frac{μ}{\sqrt{μ^{2} + \tan^{2} φ}})

\frac{\tan φ}{\sqrt{μ^{2} + \tan^{2} φ}} + \frac{μ}{1 + \cos φ} (1 - \frac{μ}{\sqrt{μ^{2} + \tan^{2} φ}}) (\sin φ + \frac{1}{\tan \frac{θ}{2}}) = \frac{μ}{\tan θ} + \sqrt{1 + {(\frac{μ}{\tan θ})}^{2}} \dots (I)

s i m i l a r l y

\frac{r}{b} = \frac{1}{μ (1 + \cos δ)} (1 - \frac{1}{\sqrt{1 + μ^{2} \tan^{2} δ}})

\frac{μ \tan δ}{\sqrt{1 + μ^{2} \tan^{2} δ}} + \frac{1}{μ (1 + \cos δ)} (1 - \frac{1}{\sqrt{1 + μ^{2} \tan^{2} δ}}) (\sin δ + \frac{1}{\tan (\frac{π}{4} - \frac{θ}{2})}) = \frac{\tan θ}{μ} + \sqrt{1 + {(\frac{\tan θ}{μ})}^{2}} \dots (I I)

\frac{1}{μ (1 + \cos δ)} (1 - \frac{1}{\sqrt{1 + μ^{2} \tan^{2} δ}}) = \frac{1}{1 + \cos φ} (1 - \frac{μ}{\sqrt{μ^{2} + \tan^{2} φ}}) \dots (I I I)

t h r e e e q n . f o r t h r e e u n k n o w n s : θ, φ, δ .

\dots \dots

Answered by ajfour last updated on 04/Feb/21

Commented by mr W last updated on 05/Feb/21

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

t o t w o e q u a t i o n s w i t h t w o u n k n o w n s .

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