Question-28241

Question Number 28241 by ajfour last updated on 22/Jan/18

Commented by ajfour last updated on 22/Jan/18

Q . 28185 (a l t e r n a t e s o l u t i o n)

Answered by ajfour last updated on 22/Jan/18

x^{2} + y^{2} - 6 x + 6 y - 16 = 0

\Rightarrow {(x - 3)}^{2} + {(y + 3)}^{2} = 34

s l o p e o f M N i s m = 4

x_{M} = 3 + \sqrt{34} \times \frac{1}{\sqrt{17}} = 3 + \sqrt{2}

y_{M} = - 3 + \sqrt{34} \times \frac{4}{\sqrt{17}} = - 3 + 4 \sqrt{2}

F o r p o i n t A w e s o l v e y = x w i t h

e q . o f A M a n d s o x_{A} = y_{A} = α i s

α - 4 \sqrt{2} + 3 = - (α - 3 - \sqrt{2})

\Rightarrow α = \frac{5}{\sqrt{2}}; h e n c e A (\frac{5}{\sqrt{2}}, \frac{5}{\sqrt{2}})

x_{N} = 3 - \sqrt{34} \times \frac{1}{\sqrt{17}} = 3 - \sqrt{2}

y_{N} = - 3 - \sqrt{34} \times \frac{4}{\sqrt{34}} = - 3 - 4 \sqrt{2}

x_{B} = y_{B} = β w h i c h w e g e t b y

s o l v i n g y = x w i t h e q . o f B N :

β + 3 + 4 \sqrt{2} = - (β - 3 + \sqrt{2})

\Rightarrow β = - \frac{5}{\sqrt{2}}; h e n c e B (- \frac{5}{\sqrt{2}}, \frac{5}{\sqrt{2}})

c e n t e r o f e l l i p s e i s m i d p o i n t o f

s e g m e n t A B, s o (0, 0) .

a n d s e m i m a j o r a x i s l e n g t h = 5

l e t u s t a k e a l o n g A B (y = x) o u r x a x i s

a n d y = - x a s n e w y a x i s

(b o t h d o w n w a r d s + v e)

t h e n e q . o f e l l i p s e i s

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, w i t h a = 5

s l o p e o f y = 4 x - 15 w i l l b e c o m e

m = \frac{4 - \tan π / 4}{1 + 4 \tan π / 4} = \frac{3}{5}

p o i n t (5, 5) w i l l b e c o m e (- 5 \sqrt{2}, 0)

t h u s e q . y = 4 x - 15 c h a n g e s t o

y = \frac{3}{5} x + 5 \sqrt{2}

a s t h i s i s t o b e t a n g e n t t o e l l i p s e

c^{2} = a^{2} m^{2} + b^{2}

\Rightarrow {(5 \sqrt{2})}^{2} = (25) {(\frac{3}{5})}^{2} + b^{2}

b^{2} = 9

e^{2} = 1 - \frac{b^{2}}{a^{2}} = 1 - \frac{9}{25}

e = \frac{4}{5}

l e n g t h o f l a t u s r e c t u m

l = \frac{2 b^{2}}{a} = \frac{2 \times 9}{5} = \frac{18}{5}

w i t h \frac{a}{e} = \frac{5}{4 / 5} = \frac{25}{4}

e q . o f d i r e c t r i c e s p a s s t h r o u g h

(\frac{25}{4 \sqrt{2}}, \frac{25}{4 \sqrt{2}}) a n d (\frac{- 25}{4 \sqrt{2}}, \frac{- 25}{4 \sqrt{2}})

a n d a r e p e r p e n d i c u l a r t o y = x

S o, e q s . a r e

y - \frac{25}{4 \sqrt{2}} = - (x - \frac{25}{4 \sqrt{2}})

\Rightarrow x + y = \pm \frac{25}{2 \sqrt{2}} .

Commented by ajfour last updated on 22/Jan/18

M N i s t a n g e n t t o t h i s e l l i p s e

a n d t h i s i s g i v e n i n q u e s t i o n, w e

c a n n o t a m e n d t h a t .

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

r a t h e r f i n d b o r h e n c e t h e

e c c e n t i c i t y .

Commented by mrW2 last updated on 22/Jan/18

I a s s u m e d t h a t t h e t a n g e n t s f r o m M

a n d N s h o u l d a t e n d s o f b o t h a x e s o f

t h e e l l i p s e . T h i s i s m u c h m o r e

c o m p l i c a t e d t h a n w h a t i s a s k e d .

I h a d t o r e a d t h e q u e s t i o n m o r e

c a r e f u l l y .

Commented by ajfour last updated on 22/Jan/18

m u s t b e, l e t m e s e e i f i c a n

f o l l o w a l o n g t h a t i n t e r p r e t a t i o n . .

Commented by mrW2 last updated on 23/Jan/18

I s t h e r e a n e a s y w a y t o g e t

c^{2} = a^{2} m^{2} + b^{2} ?

I k n o w t h i s i s t r u e . B u t I c a n d e r i v e

i t o n l y t h r o u g h a l e n g t h y w a y :

(\dots) x^{2} + (\dots) x + (\dots) = 0

Δ = {(\dots)}^{2} - 4 (\dots) (\dots) = 0

\Rightarrow c^{2} = a^{2} m^{2} + b^{2}

Commented by mrW2 last updated on 23/Jan/18

I g o t f o l l o w i n g :

r o t a t e d w i t h a n g l e - \frac{π}{4} e q . y = 4 x - 15 c h a n g e s t o

\frac{x + y}{\sqrt{2}} = \frac{4 (x - y)}{\sqrt{2}} - 15

\Rightarrow 5 y = 3 x - 15 \sqrt{2}

\Rightarrow y = \frac{3}{5} x - 3 \sqrt{2} (b u t y o u g o t y = \frac{3}{5} x + 5 \sqrt{2})

{(- 3 \sqrt{2})}^{2} = {(\frac{3}{5} \times 5)}^{2} + b^{2}

\Rightarrow b^{2} = 9 (y o u g o t a l s o b^{2} = 9)

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