If ax^2 +by^2 +2hxy+2gx+2fy+c=0 be the equation of an ellipse, find coordinates of its centre.


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Question Number 45506 by ajfour last updated on 13/Oct/18

$I f {a x}^{2} + {b y}^{2} + 2 h x y + 2 g x + 2 f y + c = 0$ $b e t h e e q u a t i o n o f a n e l l i p s e, f i n d$ $c o o r d i n a t e s o f i t s c e n t r e .$
	Answered by MrW3 last updated on 14/Oct/18
	$let θ=rotation angle of ellipse and (p,q)=center of ellipse eqn. of ellipse in uv−system is (u^2 /U^2 )+(v^2 /V^2 )=1 x=u cos θ−v sin θ+p y=u sin θ+v cos θ+q ax^2 =a cos^2 θ u^2 +a sin^2 θ v^2 −a sin 2θ uv+2ap cos θ u−2ap sin θ v+ap^2 by^2 =b sin^2 θ u^2 +b cos^2 θ v^2 +b sin 2θ uv+2bq sin θ u+2bq cos θ v+bq^2 2hxy=h sin 2θ u^2 −h sin 2θ v^2 +2h cos 2θ uv+2h(p sin θ+q cos θ) u+2h(p cos θ−q sin θ)v+2hpq 2gx=2g cos θ u−2g sin θ v+2gp 2fy=2f sin θ u+2f cos θ v+2fq+c term u^2 = A=a cos^2 θ+b sin^2 θ+h sin 2θ term v^2 = B=a sin^2 θ+b cos^2 θ−h sin 2θ term uv=C=−(a−b) sin 2θ+2h cos 2θ term u=D=2ap cos θ+2bq sin θ+2h(p sin θ+q cos θ)+2g cos θ+2f sin θ term v=E=−2ap sin θ+2bq cos θ+2h(p cos θ−q sin θ)−2g sin θ+2f cos θ term const=F=ap^2 +bq^2 +2hpq+2gp+2fq (u^2 /U^2 )+(v^2 /V^2 )−1=0 C=−a sin 2θ+b sin 2θ+2h cos 2θ=0 ⇒tan 2θ=((2h)/(a−b))⇒θ=(1/2)tan^(−1) ((2h)/(a−b)) D=2ap cos θ+2bq sin θ+2h(p sin θ+q cos θ)+2g cos θ+2f sin θ=0 ⇒(a cos θ+h sin θ)p+(b sin θ+h cos θ)q+(g cos θ+f sin θ)=0 E=−2ap sin θ+2bq cos θ+2h(p cos θ−q sin θ)−2g sin θ+2f cos θ=0 ⇒(−a sin θ+h cos θ)p+(b cos θ−h sin θ)q−g sin θ+f cos θ=0 {(a cos θ+h sin θ)(b cos θ−h sin θ)−(−a sin θ+h cos θ)(b sin θ+h cos θ)}p+{(g cos θ+f sin θ)(b cos θ−h sin θ)−(−g sin θ+f cos θ)(b sin θ+h cos θ)}=0 {(ab cos^2 θ+((bh)/2) sin 2θ−((ah)/2) sin 2θ−h^2 sin^2 θ)−(−ab sin^2 θ+((bh)/2) sin 2θ−((ah)/2) sin 2θ+h^2 cos^2 θ)}p+{(bg cos^2 θ+((bf)/2) sin 2θ−((gh)/2) sin 2θ−fh sin^2 θ)−(−bg sin^2 θ+((bf)/2) sin 2θ−((gh)/2) sin 2θ+fh cos^2 θ)}=0 (ab−h^2 )p+(bg−fh)=0 ⇒p=((fh−bg)/(ab−h^2 )) to be continued..$
	$l e t θ = r o t a t i o n a n g l e o f e l l i p s e a n d$ $(p, q) = c e n t e r o f e l l i p s e$ $e q n . o f e l l i p s e i n u v - s y s t e m i s$ $\frac{u^{2}}{U^{2}} + \frac{v^{2}}{V^{2}} = 1$ $x = u \cos θ - v \sin θ + p$ $y = u \sin θ + v \cos θ + q$ ${a x}^{2} = a \cos^{2} θ u^{2} + a \sin^{2} θ v^{2} - a \sin 2 θ u v + 2 a p \cos θ u - 2 a p \sin θ v + {a p}^{2}$ ${b y}^{2} = b \sin^{2} θ u^{2} + b \cos^{2} θ v^{2} + b \sin 2 θ u v + 2 b q \sin θ u + 2 b q \cos θ v + {b q}^{2}$ $2 h x y = h \sin 2 θ u^{2} - h \sin 2 θ v^{2} + 2 h \cos 2 θ u v + 2 h (p \sin θ + q \cos θ) u + 2 h (p \cos θ - q \sin θ) v + 2 h p q$ $2 g x = 2 g \cos θ u - 2 g \sin θ v + 2 g p$ $2 f y = 2 f \sin θ u + 2 f \cos θ v + 2 f q + c$ $t e r m u^{2} = A = a \cos^{2} θ + b \sin^{2} θ + h \sin 2 θ$ $t e r m v^{2} = B = a \sin^{2} θ + b \cos^{2} θ - h \sin 2 θ$ $t e r m u v = C = - (a - b) \sin 2 θ + 2 h \cos 2 θ$ $t e r m u = D = 2 a p \cos θ + 2 b q \sin θ + 2 h (p \sin θ + q \cos θ) + 2 g \cos θ + 2 f \sin θ$ $t e r m v = E = - 2 a p \sin θ + 2 b q \cos θ + 2 h (p \cos θ - q \sin θ) - 2 g \sin θ + 2 f \cos θ$ $t e r m c o n s t = F = {a p}^{2} + {b q}^{2} + 2 h p q + 2 g p + 2 f q$ $\frac{u^{2}}{U^{2}} + \frac{v^{2}}{V^{2}} - 1 = 0$ $C = - a \sin 2 θ + b \sin 2 θ + 2 h \cos 2 θ = 0$ $\Rightarrow \tan 2 θ = \frac{2 h}{a - b} \Rightarrow θ = \frac{1}{2} \tan^{- 1} \frac{2 h}{a - b}$ $D = 2 a p \cos θ + 2 b q \sin θ + 2 h (p \sin θ + q \cos θ) + 2 g \cos θ + 2 f \sin θ = 0$ $\Rightarrow (a \cos θ + h \sin θ) p + (b \sin θ + h \cos θ) q + (g \cos θ + f \sin θ) = 0$ $E = - 2 a p \sin θ + 2 b q \cos θ + 2 h (p \cos θ - q \sin θ) - 2 g \sin θ + 2 f \cos θ = 0$ $\Rightarrow (- a \sin θ + h \cos θ) p + (b \cos θ - h \sin θ) q - g \sin θ + f \cos θ = 0$ ${(a \cos θ + h \sin θ) (b \cos θ - h \sin θ) - (- a \sin θ + h \cos θ) (b \sin θ + h \cos θ)} p + {(g \cos θ + f \sin θ) (b \cos θ - h \sin θ) - (- g \sin θ + f \cos θ) (b \sin θ + h \cos θ)} = 0$ ${(a b \cos^{2} θ + \frac{b h}{2} \sin 2 θ - \frac{a h}{2} \sin 2 θ - h^{2} \sin^{2} θ) - (- a b \sin^{2} θ + \frac{b h}{2} \sin 2 θ - \frac{a h}{2} \sin 2 θ + h^{2} \cos^{2} θ)} p + {(b g \cos^{2} θ + \frac{b f}{2} \sin 2 θ - \frac{g h}{2} \sin 2 θ - f h \sin^{2} θ) - (- b g \sin^{2} θ + \frac{b f}{2} \sin 2 θ - \frac{g h}{2} \sin 2 θ + f h \cos^{2} θ)} = 0$ $(a b - h^{2}) p + (b g - f h) = 0$ $\Rightarrow p = \frac{f h - b g}{a b - h^{2}}$ $t o b e c o n t i n u e d . .$
	Commented by MrW3 last updated on 14/Oct/18
	$continue... {(b sin θ+h cos θ)(−a sin θ+h cos θ)−(b cos θ−h sin θ)(a cos θ+h sin θ)}q+{(g cos θ+f sin θ)(−a sin θ+h cos θ)−(−g sin θ+f cos θ)(a cos θ+h sin θ)}=0 (h^2 −ab)q+(gh−af)=0 ⇒p=((fh−bg)/(ab−h^2 )) ⇒q=((gh−af)/(ab−h^2 )) ⇒θ=(1/2)tan^(−1) ((2h)/(a−b)) A=a cos^2 θ+b sin^2 θ+h sin 2θ B=a sin^2 θ+b cos^2 θ−h sin 2θ F=ap^2 +bq^2 +2hpq+2gp+2fq+c major and minor axis of ellipse: U=(√(−(F/A))) V=(√(−(F/B)))$
	$c o n t i n u e . . .$ ${(b \sin θ + h \cos θ) (- a \sin θ + h \cos θ) - (b \cos θ - h \sin θ) (a \cos θ + h \sin θ)} q + {(g \cos θ + f \sin θ) (- a \sin θ + h \cos θ) - (- g \sin θ + f \cos θ) (a \cos θ + h \sin θ)} = 0$ $(h^{2} - a b) q + (g h - a f) = 0$ $\Rightarrow p = \frac{f h - b g}{a b - h^{2}}$ $\Rightarrow q = \frac{g h - a f}{a b - h^{2}}$ $\Rightarrow θ = \frac{1}{2} \tan^{- 1} \frac{2 h}{a - b}$ $A = a \cos^{2} θ + b \sin^{2} θ + h \sin 2 θ$ $B = a \sin^{2} θ + b \cos^{2} θ - h \sin 2 θ$ $F = {a p}^{2} + {b q}^{2} + 2 h p q + 2 g p + 2 f q + c$ $m a j o r a n d m i n o r a x i s o f e l l i p s e :$ $U = \sqrt{- \frac{F}{A}}$ $V = \sqrt{- \frac{F}{B}}$
	Commented by ajfour last updated on 14/Oct/18

	$T h a n k y o u S i r, i b e l i e v e i t c a n$ $b e f o u n d s o m e o t h e r w a y t o o, l e t$ $m e t r y . .$
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