If ax^2 +by^2 +2hxy+2gx+2fy+c=0 be the equation of an ellipse, find coordinates of center of ellipse. Q.45506 (another solution)


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Question Number 45565 by ajfour last updated on 14/Oct/18

$I f {a x}^{2} + {b y}^{2} + 2 h x y + 2 g x + 2 f y + c = 0$ $b e t h e e q u a t i o n o f a n e l l i p s e, f i n d$ $c o o r d i n a t e s o f c e n t e r o f e l l i p s e .$ $Q . 45506 (a n o t h e r s o l u t i o n)$
	Answered by ajfour last updated on 14/Oct/18
	$let centre of ellipse be (p,q). Its equation in standard orientation be (((u−p)^2 )/A^2 )+(((v−q)^2 )/B^2 )=1 when rotated by angle φ, let (u,v) → (x,y) u−p = (x−p)cos φ+(y−q)sin φ v−q = (y−q)cos φ−(x−p)sin φ Then (([(x−p)cos φ+(y−q)sin φ]^2 )/A^2 ) +(([(y−q)cos φ−(x−p)sin φ]^2 )/B^2 ) −1 = λ(ax^2 +by^2 +2hxy+2gx+2fy+c) (for all values of x and y; whether the expressions on l.h.s. and r.h.s. be equal to zero or not) let y=q_(−) , then (x−p)^2 (((cos^2 φ)/A^2 )+((sin^2 φ)/B^2 )) = λ(ax^2 +bq^2 +2hqx+2gx+2fq+c) ((coeff. of x)/(coeff. of x^2 )) = ((−2p)/1) = ((2(hq+g))/a) ⇒ ap+hq = −g ....(i) similarly if x=p_(−) , then (y−q)^2 (((sin^2 φ)/A^2 )+((cos^2 φ)/B^2 )) = λ(ap^2 +by^2 +2hpy+2gp+2fy+c) ((coeff. of y)/(coeff. of y^2 )) = ((−2q)/1) = ((2(hp+f))/b) ⇒ hp+bq = −f ....(ii) Now solving (i) & (ii) for p,q { ((ap+hq = −g )),((hp+bq = −f )) :} p = ((fh−bg)/(ab−h^2 )) ; q = ((gh−af)/(ab−h^2 )) .$
	$l e t c e n t r e o f e l l i p s e b e (p, q) .$ $I t s e q u a t i o n i n s t a n d a r d$ $o r i e n t a t i o n b e \frac{{(u - p)}^{2}}{A^{2}} + \frac{{(v - q)}^{2}}{B^{2}} = 1$ $w h e n r o t a t e d b y a n g l e ϕ,$ $l e t (u, v) \to (x, y)$ $u - p = (x - p) \cos ϕ + (y - q) \sin ϕ$ $v - q = (y - q) \cos ϕ - (x - p) \sin ϕ$ $T h e n$ $\frac{{[(x - p) \cos ϕ + (y - q) \sin ϕ]}^{2}}{A^{2}}$ $+ \frac{{[(y - q) \cos ϕ - (x - p) \sin ϕ]}^{2}}{B^{2}} - 1$ $= λ ({a x}^{2} + {b y}^{2} + 2 h x y + 2 g x + 2 f y + c)$ $(f o r a l l v a l u e s o f x a n d y; w h e t h e r$ $t h e e x p r e s s i o n s o n l . h . s . a n d r . h . s .$ $b e e q u a l t o z e r o o r n o t)$ $l e t \underline{y = q}, t h e n$ ${(x - p)}^{2} (\frac{\cos^{2} ϕ}{A^{2}} + \frac{\sin^{2} ϕ}{B^{2}})$ $= λ ({a x}^{2} + {b q}^{2} + 2 h q x + 2 g x + 2 f q + c)$ $\frac{c o e f f . o f x}{c o e f f . o f x^{2}} = \frac{- 2 p}{1} = \frac{2 (h q + g)}{a}$ $\Rightarrow a p + h q = - g . . . . (i)$ $s i m i l a r l y i f \underline{x = p}, t h e n$ ${(y - q)}^{2} (\frac{\sin^{2} ϕ}{A^{2}} + \frac{\cos^{2} ϕ}{B^{2}})$ $= λ ({a p}^{2} + {b y}^{2} + 2 h p y + 2 g p + 2 f y + c)$ $\frac{c o e f f . o f y}{c o e f f . o f y^{2}} = \frac{- 2 q}{1} = \frac{2 (h p + f)}{b}$ $\Rightarrow h p + b q = - f . . . . (i i)$ $N o w s o l v i n g (i) & (i i) f o r p, q$ ${\begin{cases} a p + h q = - g \\ h p + b q = - f \end{cases}$ $p = \frac{f h - b g}{a b - h^{2}}; q = \frac{g h - a f}{a b - h^{2}} .$
	Commented by ajfour last updated on 14/Oct/18

	$\frac{c o e f f . o f x y}{c o e f f . o f x^{2} - c o e f f . o f y^{2}}$ $= \frac{2 \sin ϕ \cos ϕ (\frac{1}{A^{2}} - \frac{1}{B^{2}})}{(\cos^{2} ϕ - \sin^{2} ϕ) (\frac{1}{A^{2}} - \frac{1}{B^{2}})}$ $= \frac{2 h}{a - b}$ $\Rightarrow \tan 2 ϕ = \frac{2 h}{a - b} .$
	Commented by MrW3 last updated on 14/Oct/18

	$t h a n k s a l o t s i r!$
	Answered by MJS last updated on 14/Oct/18

	$solving the given equation for x and y$ $x = - \frac{h y + g}{a} \pm \sqrt{. . .}$ $y = - \frac{h x + f}{b} \pm \sqrt{. . .}$ $x = - \frac{h y + g}{a} \land y = - \frac{h x + f}{b} \Rightarrow$ $\Rightarrow x = \frac{f h - b g}{a b - h^{2}} \land y = \frac{g h - a f}{a b - h^{2}}$ $no matter what kind of conic the equation is,$ $the center is (\begin{matrix} \frac{f h - b g}{a b - h^{2}} \\ \frac{g h - a f}{a b - h^{2}} \end{matrix})$
	Commented by ajfour last updated on 14/Oct/18

	$t h a n k s f o r t h e s i m p l e w a y s i r .$
	Commented by MrW3 last updated on 14/Oct/18

	$t h a n k s a l o t s i r!$
	Commented by MJS last updated on 14/Oct/18

	${you}^{'} re welcome$
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