Question-52627

Question Number 52627 by ajfour last updated on 10/Jan/19

Commented by ajfour last updated on 10/Jan/19

F i n d r a d i u s o f s u c h a c i r c l e .

Answered by mr W last updated on 10/Jan/19

Q (- h, {a h}^{2})

y_{Q}^{'} = - 2 a h

e q n . o f Q P :

y = {a h}^{2} + \frac{(x + h)}{2 a h}

y = {a x}^{2}

\Rightarrow {a h}^{2} + \frac{x + h}{2 a h} = {a x}^{2}

\Rightarrow x^{2} - \frac{x}{2 a^{2} h} - h^{2} - \frac{1}{2 a^{2}} = 0

\Rightarrow x_{P} = \frac{1}{2} [\frac{1}{2 a^{2} h} + \sqrt{\frac{1}{4 a^{4} h^{2}} + 4 (h^{2} + \frac{1}{2 a^{2}})}]

\Rightarrow x_{P} = h + \frac{1}{2 a^{2} h}

\Rightarrow y_{P} = {a h}^{2} + \frac{1}{2 a h} (2 h + \frac{1}{2 a^{2} h})

\Rightarrow y_{P} = {a h}^{2} + \frac{1}{a} + \frac{1}{4 a^{3} h^{2}}

{P Q}^{2} = {(h + \frac{1}{2 a^{2} h} + h)}^{2} + {({a h}^{2} + \frac{1}{a} + \frac{1}{4 a^{3} h^{2}} - {a h}^{2})}^{2}

{P Q}^{2} = {(2 h + \frac{1}{2 a^{2} h})}^{2} + {(\frac{1}{a} + \frac{1}{4 a^{3} h^{2}})}^{2} = y_{P}^{2}

{(2 h + \frac{1}{2 a^{2} h})}^{2} + {(\frac{1}{a} + \frac{1}{4 a^{3} h^{2}})}^{2} = {({a h}^{2} + \frac{1}{a} + \frac{1}{4 a^{3} h^{2}})}^{2}

\Rightarrow {(2 a h + \frac{1}{2 a h})}^{2} = h^{2} a^{2} (2 + a^{2} h^{2} + \frac{1}{2 a^{2} h^{2}})

l e t λ = a h

\Rightarrow {(2 λ + \frac{1}{2 λ})}^{2} = λ^{2} (2 + λ^{2} + \frac{1}{2 λ^{2}})

\Rightarrow 4 λ^{6} - 8 λ^{4} - 6 λ^{2} - 1 = 0

\Rightarrow λ \approx 1.6159

\Rightarrow R = y_{P} = \frac{1}{a} (1 + λ^{2} + \frac{1}{4 λ^{2}}) \approx \frac{3.7069}{a}

Commented by mr W last updated on 11/Jan/19

Commented by ajfour last updated on 11/Jan/19

V e r y b r i l l i a n t S i r! T h a n k y o u .

Answered by ajfour last updated on 11/Jan/19

L e t ∠ Q P F = θ, Q (- h, {a h}^{2})

\frac{d y}{d x} ∣_{Q} = - \tan θ = - 2 a h

y_{P} = R = {a h}^{2} + R \cos θ

x_{P} = - h + R \sin θ

y_{P} = {a x}_{P}^{2}

\Rightarrow R = {a h}^{2} + R \cos θ = a {(R \sin θ - h)}^{2}

\Rightarrow \cos θ = a R \sin^{2} θ - 2 a h \sin θ

\Rightarrow \cos θ = \frac{a^{2} h^{2} \sin^{2} θ}{1 - \cos θ} - 2 a h \sin θ

l e t u s n o w u s e a h = \frac{\tan θ}{2}

\Rightarrow \cos θ = \frac{\tan^{2} θ \sin^{2} θ}{4 (1 - \cos θ)} - \sin θ \tan θ

\Rightarrow 4 (1 - \cos θ) \cos^{2} θ (\cos θ + \frac{\sin^{2} θ}{\cos θ}) = \sin^{4} θ

l e t \cos θ = λ

\Rightarrow 4 (1 - λ) λ = {(1 - λ^{2})}^{2}

\Rightarrow 4 λ = {(1 + λ)}^{2} (1 - λ)

\Rightarrow λ \approx 0.2956

R = \frac{{a h}^{2}}{1 - \cos θ} = \frac{\tan^{2} θ}{4 a (1 - \cos θ)}

= \frac{(\frac{1}{\cos^{2} θ} - 1)}{4 a (1 - \cos θ)} = \frac{(1 + \cos θ)}{4 a \cos^{2} θ}

R \approx \frac{3.707}{a} .

Commented by mr W last updated on 11/Jan/19

t h a n k y o u s i r!

c a n y o u g e t λ e x a c t l y ? {i t}^{'} s a c u b i c e q n .

Commented by ajfour last updated on 12/Jan/19

T h a n k s s i r, i m e a n t i t r i e d b u t

{c o u l d n}^{'} t m a t c h w i t h t h e c a l c u l a t e d

a n s w e r i n o n e a t t e m p t .

Commented by ajfour last updated on 11/Jan/19

n o s i r, h o w c o m e ?

Commented by mr W last updated on 11/Jan/19

4 λ = {(1 + λ)}^{2} (1 - λ)

\Rightarrow λ^{3} + λ^{2} + 3 λ - 1 = 0 (c u b i c e q n .)

l e t λ = t + δ

t^{3} + (3 δ + 1) t^{2} + (3 δ^{2} + 2 δ + 3) t + δ^{3} + δ^{2} + 3 δ - 1 = 0

l e t 3 δ + 1 = 0 \Rightarrow δ = - \frac{1}{3}

\Rightarrow t^{3} + \frac{8}{3} t - \frac{52}{27} = 0

{(u + v)}^{3} - 3 u v (u + v) - (u^{3} + v^{3}) = 0

l e t t = u + v

- 3 u v = \frac{8}{3}

\Rightarrow u^{3} v^{3} = - \frac{512}{729}

\Rightarrow u^{3} + v^{3} = \frac{52}{27}

u^{3} a n d v^{3} a r e r o o t s o f

x^{2} - \frac{52}{27} x - \frac{512}{729} = 0

x = \frac{2 (13 \pm 3 \sqrt{33})}{27} = u^{3} a n d v^{3}

\Rightarrow u = \frac{\sqrt[3]{2 (13 + 3 \sqrt{33})}}{3}

\Rightarrow v = \frac{\sqrt[3]{2 (13 - 3 \sqrt{33})}}{3}

\Rightarrow λ = t + δ = u + v + δ

\Rightarrow λ = \frac{\sqrt[3]{2 (13 + 3 \sqrt{33})} + \sqrt[3]{2 (13 - 3 \sqrt{33})} - 1}{3} \approx 0.295598

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