-2
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EDIT: Thank you very much everyone for answering, and for your help. I see that I did misinterpret the equation, and missed the fact that if we have minus between two parts of the left side of the equation, it is indeed an equation of the hyperbola, not the ellipse. I apologize for this mistake, but I am just learning.


Here is the equation of the ellipse from the Precalculus book: $$\frac{(x^2)}{4} - \frac{(y - 2)}{9} = 1$$

Given the basic ellipse formula, the center lies at $$(0, 2)$$

and $$ a = \sqrt 4 = \pm 2; b = \pm3$$

The ellipse thus has its major axis along the $y$-axis, and thus vertices are:

(1) To get vertices on the major axis we move $2$ up and down along $y$-axis from the value $y = 2$ (this is the $y$ value of the center); and thus we get two vertices on the major axis: $$ (0,4) \text{ and } (0,0)$$

But in the book they say that vertices are $(\pm2, 2)$ as below

Reduces to $$18(x')^2-8(y')^2+32y'-104=0$$ or $$\frac{(x')^2}{4}-\frac{(y'-2)^2}9=1$$

The latter is the equation of a hyperbola centred at the $x'y'$-coordinates $(0,2)$ opening in the $x'$ direction with vertices $(\pm2,2)$ (in $x'y'$-coordinates) and asymptotes $y'=\pm\frac 32x'+2$

(2) to get vertices on the minor axis, we move $3$ values left and right from $0$; thus getting $$(-3,0) \text{ and } (3,0)$$

Please, help me to understand my mistake, or misunderstanding. Thank you very much!

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    $\begingroup$ That is the equation of an hyperbola, not of an ellipse. $\endgroup$ – Julián Aguirre Jul 4 '17 at 10:55
  • $\begingroup$ Not be ?$$\frac{(x^2)}{4} + \frac{(y - 2)^2}{9} = 1$$ $\endgroup$ – Khosrotash Jul 4 '17 at 10:56
  • $\begingroup$ Neither the equation in the question, nor the one suggested by @Khosrotash are equations of ellipses... The original one is a parabola and the comment is a hyperbola $\endgroup$ – lioness99a Jul 4 '17 at 11:40
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    $\begingroup$ @lioness99a: Khosrotash's equation certainly looks like an ellipse to me. $\endgroup$ – TonyK Jul 4 '17 at 11:50
  • $\begingroup$ @TonyK My bad, misread the sign between the two fractions. It is, indeed an ellipse $\endgroup$ – lioness99a Jul 4 '17 at 11:54
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Hint. The center of the hyperbola is along the line $y=2$ hence, in order to finf the minor axis you should solve $$\frac{(x^2)}{4} - \frac{(2 - 2)^2}{9} = 1\Leftrightarrow x^2=4.$$

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