Question-48638

Question Number 48638 by mr W last updated on 26/Nov/18

Commented by mr W last updated on 26/Nov/18

F i n d t h e l o c u s o f P s u c h t h a t P O

i s t h e b i s e c t o r o f ∠ A P B .

Commented by ajfour last updated on 28/Nov/18

Commented by ajfour last updated on 28/Nov/18

\frac{a}{\sin ϕ} = \frac{r}{\sin (θ + ϕ)} &

\frac{b}{\sin ϕ} = \frac{r}{\sin (\frac{π}{2} - θ + ϕ)} = \frac{r}{\cos (θ - ϕ)}

\Rightarrow \sin θ \cos ϕ + \cos θ \sin ϕ = (r / a) \sin ϕ

\sin θ \sin ϕ + \cos θ \cos ϕ + = (r / b) \sin ϕ

\Rightarrow \sin θ = \frac{r \sin ϕ (\frac{\cos ϕ}{a} - \frac{\sin ϕ}{b})}{\cos^{2} ϕ - \sin^{2} ϕ}

\cos θ = \frac{r \sin ϕ (\frac{\cos ϕ}{b} - \frac{\sin ϕ}{a})}{\cos^{2} ϕ - \sin^{2} ϕ}

\Rightarrow 1 = \frac{r^{2} \sin^{2} ϕ [{(\frac{\cos ϕ}{a} - \frac{\sin ϕ}{b})}^{2} + {(\frac{\cos ϕ}{b} - \frac{\sin ϕ}{a})}^{2}]}{{(\cos^{2} ϕ - \sin^{2} ϕ)}^{2}}

\Rightarrow 1 = \frac{r^{2} \sin^{2} ϕ [\frac{1}{a^{2}} + \frac{1}{b^{2}} - \frac{4 \sin ϕ \cos ϕ}{a b}]}{{(\cos^{2} ϕ - \sin^{2} ϕ)}^{2}}

\Rightarrow r^{2} = \frac{\cos^{2} 2 ϕ}{\sin^{2} ϕ (\frac{1}{a^{2}} + \frac{1}{b^{2}} - \frac{2 \sin 2 ϕ}{a b})}

r^{2} = \frac{\cot^{2} ϕ - 1}{\frac{1}{a^{2}} + \frac{1}{b^{2}} - \frac{2 \sin 2 ϕ}{a b}}

\frac{2 r d r}{d ϕ} = 0 \Rightarrow

- 2 \cot ϕ cosec^{2} ϕ (\frac{1}{a^{2}} + \frac{1}{b^{2}} - \frac{2 \sin 2 ϕ}{a b})

= \frac{- 4 \cos 2 ϕ}{a b} (\cot^{2} ϕ - 1)

\Rightarrow \frac{a^{2} + b^{2}}{2 a b} - \sin 2 ϕ = \cos^{2} 2 ϕ \tan ϕ

l e t \tan ϕ = t, \frac{a^{2} + b^{2}}{2 a b} = c

\Rightarrow t {(\frac{1 - t^{2}}{1 + t^{2}})}^{2} + \frac{2 t}{1 + t^{2}} - c = 0

\Rightarrow t {(1 - t^{2})}^{2} + 2 t (1 + t^{2}) - c {(1 + t^{2})}^{2} = 0

\dots . .

Commented by mr W last updated on 28/Nov/18

t h a n k s s i r!

w e c a n a l s o b u i l d t h e e q n . w . r . t . θ :

\frac{r}{a} = \cos θ + \sin θ \frac{\frac{b}{a} \tan θ - 1}{\tan θ - \frac{b}{a}}

M J S s i r h a s s o l v e d t h e m a x i m u m f o r

t h i s e q n . i n Q 48739 .

Answered by ajfour last updated on 26/Nov/18

L e t P (x, y)

\cos θ = \frac{\overset{―}{A P} . \overset{―}{O P}}{∣ \overset{―}{A P} ∣∣ \overset{―}{O P} ∣} = \frac{\overset{―}{B P} . \overset{―}{O P}}{∣ \overset{―}{B P} ∣∣ \overset{―}{O P} ∣}

\frac{x (x - a) + y^{2}}{\sqrt{{(x - a)}^{2} + y^{2}} \sqrt{x^{2} + y^{2}}} = \frac{x^{2} + y (y - b)}{\sqrt{x^{2} + {(y - b)}^{2}} \sqrt{x^{2} + y^{2}}} .

Commented by ajfour last updated on 26/Nov/18

s h o u l d n o t b e o m i t t e d, i t m i g h t

s e r v e t o r e m i n d t h a t o r i g i n c a n n o t

b e a p o i n t o f l o c u s (f o r θ i s n o t

d e i n e d t h e n) .

Commented by mr W last updated on 26/Nov/18

O k s i r!

a s s u m e P i s o n l y i n f i r s t q u a d r a n t .

l e t R = O P

i s t h e r e a m a x . R ?

Commented by ajfour last updated on 26/Nov/18

i g u e s s n o R_{m a x} .

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

r \to \infty s e e m t o b e o b e y t h e e q u a t i o n .

Commented by mr W last updated on 26/Nov/18

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

∣ O P ∣ o n b o t h s i d e s c a n b e o m i t t e d . s o

t h e e q n . i s

\frac{x^{2} + y^{2} - a x}{\sqrt{{(x - a)}^{2} + y^{2}}} = \frac{x^{2} + y^{2} - b y}{\sqrt{x^{2} + {(y - b)}^{2}}}

I^{'} m t r y i n g t o i m a g e h o w i t l o o k s l i k e .

Commented by mr W last updated on 27/Nov/18

Commented by mr W last updated on 27/Nov/18

f o r t h e c a s e a > b, t h e l o c u s o f p o i n i P

i n t h e f i r s t q u a d r a n t i s t h e r e d c u r v e .

w e h a v e p o i n t P_{1} w i t h d i s t a n c e R_{1} t o

t h e o r i g i n . w e h a v e p o i n t P_{2} = A w i t h

d i s t a n c e R_{2} t o t h e o r i g i n (R_{2} = a) .

w e c a n s e e 0 < r ⩽ R_{1} a n d R_{2} < r < \infty .

t h a t m e a n s f o r R_{1} < r < R_{2} t h e r e i s

n o s o l u t i o n f o r θ . R_{1} i s a k i n d o f l o c a l

m a x i m u m f o r r .

Commented by mr W last updated on 27/Nov/18

Commented by mr W last updated on 27/Nov/18

t h i s d i a g r a m s h o w s t h e c a s e a = 4, b = 3 .

f o r r i n t h e s h a d e d a r e a t h e r e i s n o

s o l u t i o n . s i n c e r i s t h e l e n g t h o f t h e

r o d, t h a t m e a n s w e c a n n o t f i n d a

s o l u t i o n f o r s o m e l e n g t h e s o f r o d .

Commented by ajfour last updated on 27/Nov/18

Q u i t e i n t e r e s t i n g, s i r .

Commented by mr W last updated on 27/Nov/18

I f a i l e d t o d e t e r m i n e R_{1}, t h e l o c a l

m a x i m u m o f r f o r c a s e a \neq b .

Commented by ajfour last updated on 28/Nov/18

i a m n o t i n g o o d h e a l t h s i r .

Commented by mr W last updated on 28/Nov/18

s o r r y t o h e a r t h i s s i r . I w i s h y o u a f u l l

r e c o v e r y a . s . a . p .!

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