Question-10223

Question Number 10223 by amir last updated on 30/Jan/17

Answered by mrW1 last updated on 03/Feb/17

t o q u e s t i o n 1 :

x y = 1

y = \frac{1}{x}

y^{'} = - \frac{1}{x^{2}}

l e t B a p o i n t o n t h e c u r v e a n d A B i s

a o r t h o g o n a l l i n e t o t h e c u r v e .

l e t p o i n t B h a v e t h e c o o r d i n a t e s (t, s) .

s = \frac{1}{t}

t h e s l o p e o f t a n g e n t l i n e o n p o i n t B :

m_{t} = y^{'} (t) = - \frac{1}{t^{2}}

t h e s l o p e o f t h e o r t h o g o n a l l i n e :

m_{o} = - \frac{1}{m_{t}} = t^{2}

t h e e q u a t i o n o f t h e o r t h o g o n a l l i n e :

y - s = m_{o} (x - t)

y - \frac{1}{t} = t^{2} (x - t)

s i n c e A (a, a) i s a p o i n t o n i t, w e g e t

a - \frac{1}{t} = t^{2} (a - t)

t^{4} - {a t}^{3} + a t - 1 = 0

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

\Rightarrow t_{1} = - 1 (p o i n t D) o r

\Rightarrow t_{2} = 1 (p o i n t C) o r

\Rightarrow t - a t + 1 = 0

\Rightarrow\Rightarrow t_{3, 4} = \frac{a \pm \sqrt{a^{2} - 4}}{2}

i f ∣ a ∣< 2, t_{3, 4} = u n r e a l

i f ∣ a ∣= 2, t_{3, 4} = 1 = t_{2}

i f ∣ a ∣> 2, t_{3, 4} = \frac{a \pm \sqrt{a^{2} - 4}}{2} (p o i n t B a n d B^{'})

i . e . f o r a n y p o i n t A (a, a) t h e r e a r e

a t l e a s t 2 o r t h o g o n a l l i n e s (A C, A D) .

i f ∣ a ∣> 2, t h e r e a r e e v e n 2 a d d i t i o n a l

o r t h o g o n a l l i n e s (A B, {A B}^{'}) .

Commented by mrW1 last updated on 02/Feb/17

Commented by amir last updated on 02/Feb/17

t h a n k y o u s o m u c h . b u t Q n o . 2 a n d n o . 3 ?

Commented by amir last updated on 02/Feb/17

w i t h w h i c h p r o g r a m i c a n d r a w t h i s

d i a g r a m s ?

Commented by mrW1 last updated on 03/Feb/17

p l e a s e t r y t h e a p p g e o g e b r a .

Answered by mrW1 last updated on 04/Feb/17

t o q u e s t i o n 2 :

i f ∣ a ∣⩽ 2, i . e . - 2 ⩽ a ⩽ 2, t h e r e a r e o n l y

2 o r t h o g o n a l l i n e s f r o m p o i n t A :

A D w i t h s l o p e = 45 °

A C w i t h s l o p e = 45 °

t h e a n g e l b e t w e e n t h e m i s 0 ° .

i f ∣ a ∣> 2, i . e . a < - 2 o r a > 2, t h e r e a r e

4 o r t h o g o n a l l i n e s f r o m p o i n t A :

A D w i t h s l o p e = 45 °

A C w i t h s l o p e = 45 °

t h e a n g e l b e t w e e n t h e m i s 0 ° .

A B w i t h s l o p e a n g e l, l e t s s a y α_{1}

{A B}^{'} w i t h s l o p e a n g e l, l e t s s a y α_{2}

t h e a n g e l b e t w e e n t h e m i s, l e t s s a y α

α = α_{1} - α_{2}

\tan α_{1} = - \frac{1}{f^{'} (t_{3})} = t_{3}^{2} = {(\frac{a + \sqrt{a^{2} - 4}}{2})}^{2}

\tan α_{2} = - \frac{1}{f^{'} (t_{4})} = t_{4}^{2} = {(\frac{a - \sqrt{a^{2} - 4}}{2})}^{2}

\tan α = \tan (α_{1} - α_{2}) = \frac{\tan α_{1} - \tan α_{2}}{1 + \tan_{1} \tan α_{2}}

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

= 4 \times \frac{{(a + \sqrt{a^{2} - 4})}^{2} - {(a - \sqrt{a^{2} - 4})}^{2}}{16 + {(a + \sqrt{a^{2} - 4})}^{2} {(a - \sqrt{a^{2} - 4})}^{2}}

= 4 \times \frac{(a + \sqrt{a^{2} - 4} + a - \sqrt{a^{2} - 4}) (a + \sqrt{a^{2} - 4} - a + \sqrt{a^{2} - 4})}{16 + {(a^{2} - a^{2} + 4)}^{2}}

= 4 \times \frac{(2 a) (2 \sqrt{a^{2} - 4})}{32}

= \frac{a \sqrt{a^{2} - 4}}{2}

α = \tan^{- 1} (\frac{a \sqrt{a^{2} - 4}}{2})

Commented by amir last updated on 04/Feb/17

t h a n k y o u f o r e v r e y t h i n g . b u t p l e a s e

c h e c k y o u r c a l c u l a t i o n . i g o t t h e a n g l e

α = {t a n}^{- 1} (\frac{1}{2} a \sqrt{a^{2} - 4})

Commented by mrW1 last updated on 04/Feb/17

y o u r a r e r i g h t . i t i s f i x e d n o w . t h a n k y o u!

Commented by mrW1 last updated on 04/Feb/17

w h a t a b o u t Q 1 a n d Q 2 i f w e h a v e p o i n t A (a, - a) ?

Commented by amir last updated on 08/Feb/17

i n c a s e o f p o i n t B (a, - a), t h e r e a r e

o n l y 2 p e r p e n d i c u l a r e l i n e s .

y = \frac{1}{x} \Rightarrow y^{^{'}} = - \frac{1}{x^{2}} \Rightarrow m_{t} = - \frac{1}{t^{2}}, m_{p} = + t^{2}

t h a t M (t, \frac{1}{t}) a n d N (s, \frac{1}{s}) a r e t h e i n t e r s e c t i o n p o i n t s

o f t h e c u r v e a n d t h e p e r . l i n e s .

o l s o m_{p} b e t h e s l o p e o f p e \underset{}{r} . l i n e

y - y_{M} = m_{p} (x - x_{M}) \Rightarrow y - \frac{1}{t} = t^{2} (x - t)

- a - \frac{1}{t} = t^{2} (a - t) \Rightarrow t^{4} - {a t}^{3} - a t - 1 = 0

(t^{2} + 1) (t^{2} - a t - 1) = 0, t^{2} + 1 \neq 0

\Rightarrow t = \frac{a \pm \sqrt{a^{2} + 4}}{2} \Rightarrow (\begin{matrix} M (\frac{a + \sqrt{a^{2} + 4}}{2}, \frac{a - \sqrt{a^{2} + 4}}{- 2}) \\ N (\frac{a - \sqrt{a^{2} + 4}}{2}, \frac{a + \sqrt{a^{2} + 4}}{- 2}) \end{matrix})

m_{B N} = \frac{- a - \frac{a + \sqrt{a^{2} + 4}}{- 2}}{a - \frac{a - \sqrt{a^{2} + 4}}{2}} = \frac{1}{4} {(a - \sqrt{a^{2} + 4})}^{2}

m_{B M} = \frac{- a - \frac{a - \sqrt{a^{2} + 4}}{- 2}}{a - \frac{a + \sqrt{a^{2} + 4}}{2}} = \frac{1}{4} {(a + \sqrt{a^{2} + 4})}^{2}

\tan α =∣ \frac{m_{B N} - m_{B M}}{1 + m_{B N} \times m_{B M}} ∣=

∣ \frac{\frac{1}{4} {(a - \sqrt{a^{2} + 4})}^{2} - \frac{1}{4} {(a + \sqrt{a^{2} + 4})}^{2}}{1 + \frac{1}{4} {(a - \sqrt{a^{2} + 4})}^{2} \times \frac{1}{4} {(a + \sqrt{a^{2} + 4})}^{2}} ∣= \frac{1}{2} a \sqrt{a^{2} + 4}

α = \tan^{- 1} (\frac{1}{2} a \sqrt{a^{2} + 4})

Commented by amir last updated on 07/Feb/17

Commented by amir last updated on 07/Feb/17

d e a r m r W 1! p l e a s e d r a w a n i c e d i a g r a m

w i t h g e o g e b r a a n d p o s t i t h e r e . t n x .

Commented by mrW1 last updated on 08/Feb/17

Commented by amir last updated on 08/Feb/17

t h a n k y o u v e r y m u c h m r w 1 .

y o u a r e a r e a l m a t h m a t i c m a n .

g o o d l o c k!

Answered by mrW1 last updated on 03/Feb/17

t o q u e s t i o n 3 :

i f ∣ a ∣⩽ 2, t h e r e a r e 2 o r t h o g o n a l l i n e s

f r o m A . t h e a r e a b e t w e e n t h e m a n d

t h e c u r v e i s 0 .

i f ∣ a ∣> 2, t h e r e a r e 4 o r t h o g o n a l l i n e s

f r o m A .

F_{1} = a r e a o f {A R P B}^{'}

F_{2} = a r e a o f A R Q B

F_{3} = a r e a o f B^{'} B Q P

F = a r e a b e t w e e n o r t h o g o n a l l i n e s

a n d t h e c u r v e, {A B B}^{'} .

F = F_{1} - F_{2} - F_{3}

F_{1} = \frac{1}{2} (a + y_{4}) (a - x_{4}) = \frac{1}{2} (a + \frac{1}{t_{4}}) (a - t_{4}) = \frac{1}{2} (a + \frac{2}{a - \sqrt{a^{2} - 4}}) (a + \frac{a - \sqrt{a^{2} - 4}}{2})

F_{2} = \frac{1}{2} (a + y_{3}) (a - x_{3}) = \frac{1}{2} (a + \frac{1}{t_{3}}) (a - t_{3}) = \frac{1}{2} (a + \frac{2}{a + \sqrt{a^{2} - 4}}) (a + \frac{a + \sqrt{a^{2} - 4}}{2})

F_{3} = \int_{x_{4}}^{x_{3}} \frac{1}{x} d x = (\ln ∣ x ∣) ∣_{t_{4}}^{t_{3}} = \ln ∣ \frac{t_{3}}{t_{4}} ∣= \ln ∣ \frac{a + \sqrt{a^{2} - 4}}{a - \sqrt{a^{2} - 4}} ∣

F = \frac{1}{2} (a + \frac{2}{a - \sqrt{a^{2} - 4}}) (a + \frac{a - \sqrt{a^{2} - 4}}{2}) - \frac{1}{2} (a + \frac{2}{a + \sqrt{a^{2} - 4}}) (a + \frac{a + \sqrt{a^{2} - 4}}{2}) - \ln ∣ \frac{a + \sqrt{a^{2} - 4}}{a - \sqrt{a^{2} - 4}} ∣

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