Find-the-symmetry-axes-if-any-of-the-following-curve-x-2-y-2-2xy-2x-8y-2-0-

Question Number 85298 by mr W last updated on 20/Mar/20

F i n d t h e s y m m e t r y a x e s (i f a n y) o f

t h e f o l l o w i n g c u r v e :

x^{2} + y^{2} - 2 x y + 2 x - 8 y - 2 = 0

Commented by jagoll last updated on 20/Mar/20

x^{2} + 2 x + 1 + y^{2} - 8 y + 16 = 2 x y + 19

{(x + 1)}^{2} + {(y - 4)}^{2} = 2 x y + 19

n e x t

Commented by mr W last updated on 20/Mar/20

a n y r e s u l t s i r ?

Answered by MJS last updated on 21/Mar/20

y = x + 4 \pm \sqrt{6 (x + 3)} defined for x ⩾ - 3

x = y - 1 \pm \sqrt{3 (2 y + 1)} defined for y ⩾ - \frac{1}{2}

\Rightarrow {it}^{'} s a parabola with direction x^{+} and y^{+}

\dots under construction \dots

Commented by mr W last updated on 22/Mar/20

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

i g o t t h e s y m m e t r y a x i s y = x + \frac{5}{2} a n d

h a v e c h e c k e d g r a p h i c a l l y .

Commented by mr W last updated on 21/Mar/20

Answered by mr W last updated on 21/Mar/20

M e t h o d 1

x^{2} + y^{2} - 2 x y + 2 x - 8 y - 2 = 0

\Rightarrow {(x - y + 1)}^{2} - 6 y - 3 = 0

i t i s t o s e e t h a t t h e c u r v e i s a p a r a b o l a,

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

a x i s .

t h e s y m m e t r y a x i s o f a p a r a b o l a i s

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

(n o t t a n g e n t) t h e p a r a b o l a a t o n l y o n e

p o i n t .

{l e t}^{'} s s a y t h i s s y m m e t r y a x i s i s

x = m y + k, o r y = \frac{x}{m} - \frac{k}{m} .

t h e i n t e r s e c t i o n o f b o t h i s t h e

s o l u t i o n o f

{\begin{cases} {(x - y + 1)}^{2} - 6 y - 3 = 0 \\ x = m y + k \end{cases}

{(m y + k - y + 1)}^{2} - 6 y - 3 = 0

{[(m - 1) y + (k + 1)]}^{2} - 6 y - 3 = 0

{(m - 1)}^{2} y^{2} + 2 [(m - 1) (k + 1) - 3] y + {(k + 1)}^{2} - 3 = 0

s i n c e i t s h o u l d a l w a y s h a v e o n l y o n e

r o o t, t h e y^{2} t e r m m u s t v a n i s h, i . e .

m = 1 . t h e e q n . i s t h e n

- 6 y + {(k + 1)}^{2} - 3 = 0

\Rightarrow y = \frac{{(k + 1)}^{2} - 3}{6}

\Rightarrow x = \frac{{(k + 1)}^{2} - 3}{6} + k

s u c h t h a t t h i s l i n e i s {p a r a b o l a}^{'} s a x i s,

t h e i n t e r s e c t i o n p o i n t i s t h e v e r t e x,

i . e . v e r t e x i s (\frac{{(k + 1)}^{2} - 3}{6} + k, \frac{{(k + 1)}^{2} - 3}{6}) .

w e k n o w a t v e r t e x t h e t a n g e n t o f t h e

p a r a b o l a i s p e r p e n d i c u l a r t o t h e

(s y m m e t r y) a x i s o f t h e p a r a b o l a,

i . e . y^{'} = - m = - 1 .

f r o m {(x - y + 1)}^{2} - 6 y - 3 = 0 w e g e t

2 (x - y + 1) (1 - y^{'}) - 6 y^{'} = 0

a t t h e v e r t e x :

2 (k + 1) (1 + 1) + 6 = 0

\Rightarrow k = - \frac{5}{2}

t h e r e f o r e t h e s y m m e t r y a x i s i s

x = m y + k = y - \frac{5}{2}

o r y = x + \frac{5}{2}

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