Given (Γ): x^4 −16(y^2 −2y)^2 =0 Show that (Γ) is a reunion of and Ellipsis and an hyperbole then give their equations.


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Question Number 138470 by mathocean1 last updated on 13/Apr/21

$G i v e n (Γ) : x^{4} - 16 {(y^{2} - 2 y)}^{2} = 0$ $S h o w t h a t (Γ) i s a r e u n i o n o f a n d$ $E l l i p s i s a n d a n h y p e r b o l e t h e n g i v e$ $t h e i r e q u a t i o n s .$
	Answered by MJS_new last updated on 14/Apr/21

	$x^{4} - 16 {(y^{2} - 2 y)}^{2} = 0$ $(x^{2} - 4 (y^{2} - 2 y)) (x^{2} + 4 (y^{2} - 2 y)) = 0$ $\Rightarrow$ $(x^{2} - 4 (y^{2} - 2 y)) = 0 \lor (x^{2} + 4 (y^{2} - 2 y)) = 0$ $x^{2} - 4 (y^{2} - 2 y) = 0 \Leftrightarrow y = 1 \pm \frac{\sqrt{x^{2} + 4}}{2} hyperbola$ $x^{2} + 4 (y^{2} - 2 y) = 0 \Leftrightarrow y = 1 \pm \frac{\sqrt{4 - x^{2}}}{2} ellipse$
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