find the center point ellips 5x^2 +8y^2 −24x−24y+4xy=0


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Question Number 78119 by jagoll last updated on 14/Jan/20

$f i n d t h e c e n t e r p o i n t e l l i p s$ $5 x^{2} + 8 y^{2} - 24 x - 24 y + 4 x y = 0$
Commented by jagoll last updated on 14/Jan/20

$t h i s p r o b l e m i s r i g h t o r n o t ?$
Commented by mr W last updated on 14/Jan/20

$q u e s t i o n i s r i g h t a n d g o o d .$
Commented by MJS last updated on 14/Jan/20

$solving for x :$ $x = - \frac{2}{5} y + \frac{12}{5} \pm \frac{6}{5} \sqrt{- y^{2} + 2 y + 4}$ $this is defined for 1 - \sqrt{5} ⩽ y ⩽ 1 + \sqrt{5} \Rightarrow$ $\Rightarrow y - coordinate of center = 1$ $solving for y :$ $y = - \frac{x}{4} + \frac{3}{2} \pm \frac{3}{4} \sqrt{- x^{2} + 4 x l + 4}$ $this is defined for 2 - 2 \sqrt{2} ⩽ x ⩽ 2 + 2 \sqrt{2} \Rightarrow$ $\Rightarrow x - coordinate of center = 2$
Commented by jagoll last updated on 14/Jan/20

$t h a n k s s i r$
Commented by jagoll last updated on 14/Jan/20

$f i n a l l y t h e c e n t e r p o i n t a t (2, 1)$
Commented by MJS last updated on 14/Jan/20

$yes$
	Answered by mr W last updated on 15/Jan/20

	$i f y o u {d o n}^{'} t l i k e t o l e a r n t o o m a n y$ $f o r m u l a s, y o u c a n d e t e r m i n e t h e$ $p a r a m e t e r s o f t h e e l l i p s e i n f o l l o w i n g$ $w a y :$ $s t e p 1 : d e t e r m i n e c e n t e r p o i n t$ $t h a t m e a n s t o d e t e r m i n e t h e r a n g e o f$ $x a n d y o r d i n a t e s . (s e e a l s o s i r {M J S}^{'} p o s t)$ $w e f o r m e q n . a s q u a d r a t i c e q n . w . r . t . y :$ $8 y^{2} + 4 (x - 6) y + 5 x^{2} - 24 x = 0$ $Δ = 4^{2} {(x - 6)}^{2} - 4 \times 8 \times (5 x^{2} - 24 x) ⩾ 0$ ${(x - 6)}^{2} - 2 (5 x^{2} - 24 x) ⩾ 0$ $- x^{2} + 4 x + 4 ⩾ 0$ $x^{2} - 4 x - 4 ⩽ 0$ ${(x - 2)}^{2} - 8 ⩽ 0$ $\Rightarrow 2 - 2 \sqrt{2} ⩽ x ⩽ 2 + 2 \sqrt{2}$ $\Rightarrow x_{C} = 2$ $w e f o r m e q n . a s q u a d r a t i c e q n . w . r . t . x :$ $5 x^{2} + 4 (y - 6) x + 8 y^{2} - 24 y = 0$ $Δ = 4^{2} {(y - 6)}^{2} - 4 \times 5 (8 y^{2} - 24 y) ⩾ 0$ ${(y - 6)}^{2} - 5 (2 y^{2} - 6 y) ⩾ 0$ $- y^{2} + 2 y + 4 ⩾ 0$ $y^{2} - 2 y - 4 ⩽ 0$ ${(y - 1)}^{2} - 5 ⩽ 0$ $\Rightarrow 1 - \sqrt{5} ⩽ y ⩽ 1 + \sqrt{5}$ $\Rightarrow y_{C} = 1$ $\Rightarrow c e n t e r p o i n t o f e l l i p s e i s C (2, 1)$ $s t e p 2 : m a k e t h e c e n t e r p o i n t a s o r i g i n$ $r e p l a c e x w i t h x + x_{C} a n d y w i t h y + y_{C}$ $i n t h e e q n .$ $5 {(x + 2)}^{2} + 8 {(y + 1)}^{2} - 24 (x + 2) - 24 (y + 1) + 4 (x + 2) (y + 1) = 0$ $\Rightarrow 5 x^{2} + 8 y^{2} + 4 x y - 36 = 0$ $s t e p 3 : f i n d t h e m a j o r a n d m i n o r a x i s$ $e x p r e s s t h e e q n . i n p o l a r f o r m w i t h$ $x = r \cos θ, y = r \sin θ$ $5 r^{2} \cos^{2} θ + 8 r^{2} \sin^{2} θ + 4 r^{2} \sin θ \cos θ - 36 = 0$ $[\frac{5}{2} (\cos 2 θ + 1) + 4 (1 - \cos 2 θ) + 2 \sin 2 θ] r^{2} - 36 = 0$ $(13 - 3 \cos 2 θ + 4 \sin 2 θ) r^{2} = 72$ $[13 + 5 (- \frac{3}{5} \cos 2 θ + \frac{4}{5} \sin 2 θ)] r^{2} = 72$ $[13 + 5 (- \sin α \cos 2 θ + \cos α \sin 2 θ)] r^{2} = 72$ $[13 + 5 \sin (2 θ - α)] r^{2} = 72$ $\Rightarrow r = \frac{6 \sqrt{2}}{\sqrt{13 + 5 \sin (2 θ - α)}}$ $r_{m a x} = m a j o r a x i s = a = \frac{6 \sqrt{2}}{\sqrt{13 - 5}} = 3$ $a t 2 θ - α = - \frac{π}{2} o r θ = \tan^{- 1} \frac{1}{3} - \frac{π}{4}$ $r_{m i n} = m i n o r a x i s = b = \frac{6 \sqrt{2}}{\sqrt{13 + 5}} = 2$ $a t 2 θ - α = \frac{π}{2} o r θ = \tan^{- 1} \frac{1}{3} + \frac{π}{4}$ $w e k n o w i n t h i s w a y a l s o t h a t t h e$ $e l l i p s e i s r o t a t e d b y a n g l e \tan^{- 1} \frac{1}{3} - \frac{π}{4}$
	Commented by mr W last updated on 15/Jan/20

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