Find the symmetry axes (if any) of the following curve: x^2 +y^2 −2xy+2x−8y−2=0


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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

	$y = x + 4 \pm \sqrt{6 (x + 3)} defined for x ⩾ - 3$ $x = y - 1 \pm \sqrt{3 (2 y + 1)} defined for y ⩾ - \frac{1}{2}$ $\Rightarrow {it}^{'} s a parabola with direction x^{+} and y^{+}$ $. . . under construction . . .$
	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
	$Method 1 x^2 +y^2 −2xy+2x−8y−2=0 ⇒(x−y+1)^2 −6y−3=0 it is to see that the curve is a parabola, which has one and only one symmetry axis. the symmetry axis of a parabola is one of the lines which intersect (not tangent) the parabola at only one point. let′s say this symmetry axis is x=my+k, or y=(x/m)−(k/m). the intersection of both is the solution of { (((x−y+1)^2 −6y−3=0)),((x=my+k)) :} (my+k−y+1)^2 −6y−3=0 [(m−1)y+(k+1)]^2 −6y−3=0 (m−1)^2 y^2 +2[(m−1)(k+1)−3]y+(k+1)^2 −3=0 since it should always have only one root, the y^2 term must vanish, i.e. m=1. the eqn. is then −6y+(k+1)^2 −3=0 ⇒y=(((k+1)^2 −3)/6) ⇒x=(((k+1)^2 −3)/6)+k such that this line is parabola′s axis, the intersection point is the vertex, i.e. vertex is ((((k+1)^2 −3)/6)+k, (((k+1)^2 −3)/6)). we know at vertex the tangent of the parabola is perpendicular to the (symmetry) axis of the parabola, i.e. y′=−m=−1. from (x−y+1)^2 −6y−3=0 we get 2(x−y+1)(1−y′)−6y′=0 at the vertex: 2(k+1)(1+1)+6=0 ⇒k=−(5/2) therefore the symmetry axis is x=my+k=y−(5/2) or y=x+(5/2)$
	$M e t h o d 1$ $x^{2} + y^{2} - 2 x y + 2 x - 8 y - 2 = 0$ $\Rightarrow {(x - y + 1)}^{2} - 6 y - 3 = 0$ $i t i s t o s e e t h a t t h e c u r v e i s a p a r a b o l a,$ $w h i c h h a s o n e a n d o n l y o n e s y m m e t r y$ $a x i s .$ $t h e s y m m e t r y a x i s o f a p a r a b o l a i s$ $o n e o f t h e l i n e s w h i c h i n t e r s e c t$ $(n o t t a n g e n t) t h e p a r a b o l a a t o n l y o n e$ $p o i n t .$ ${l e t}^{'} s s a y t h i s s y m m e t r y a x i s i s$ $x = m y + k, o r y = \frac{x}{m} - \frac{k}{m} .$ $t h e i n t e r s e c t i o n o f b o t h i s t h e$ $s o l u t i o n o f$ ${\begin{cases} {(x - y + 1)}^{2} - 6 y - 3 = 0 \\ x = m y + k \end{cases}$ ${(m y + k - y + 1)}^{2} - 6 y - 3 = 0$ ${[(m - 1) y + (k + 1)]}^{2} - 6 y - 3 = 0$ ${(m - 1)}^{2} y^{2} + 2 [(m - 1) (k + 1) - 3] y + {(k + 1)}^{2} - 3 = 0$ $s i n c e i t s h o u l d a l w a y s h a v e o n l y o n e$ $r o o t, t h e y^{2} t e r m m u s t v a n i s h, i . e .$ $m = 1 . t h e e q n . i s t h e n$ $- 6 y + {(k + 1)}^{2} - 3 = 0$ $\Rightarrow y = \frac{{(k + 1)}^{2} - 3}{6}$ $\Rightarrow x = \frac{{(k + 1)}^{2} - 3}{6} + k$ $s u c h t h a t t h i s l i n e i s {p a r a b o l a}^{'} s a x i s,$ $t h e i n t e r s e c t i o n p o i n t i s t h e v e r t e x,$ $i . e . v e r t e x i s (\frac{{(k + 1)}^{2} - 3}{6} + k, \frac{{(k + 1)}^{2} - 3}{6}) .$ $w e k n o w a t v e r t e x t h e t a n g e n t o f t h e$ $p a r a b o l a i s p e r p e n d i c u l a r t o t h e$ $(s y m m e t r y) a x i s o f t h e p a r a b o l a,$ $i . e . y^{'} = - m = - 1 .$ $f r o m {(x - y + 1)}^{2} - 6 y - 3 = 0 w e g e t$ $2 (x - y + 1) (1 - y^{'}) - 6 y^{'} = 0$ $a t t h e v e r t e x :$ $2 (k + 1) (1 + 1) + 6 = 0$ $\Rightarrow k = - \frac{5}{2}$ $t h e r e f o r e t h e s y m m e t r y a x i s i s$ $x = m y + k = y - \frac{5}{2}$ $o r y = x + \frac{5}{2}$
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