How do you write 4x^2 + y^2 – 4y – 12 = 0 in standard form and why type of conic is it?

${\left(x - 0\right)}^{2} / 4 + {\left(y - 2\right)}^{2} / 16 = 1$
Ellipse

Explanation:

by completing the square
$4 {x}^{2} + {y}^{2} - 4 y - 12 = 0$

$4 {x}^{2} + {y}^{2} - 4 y + 4 - 4 - 12 = 0$

$4 {\left(x - 0\right)}^{2} + {\left(y - 2\right)}^{2} = 16$

$\frac{4 {\left(x - 0\right)}^{2}}{16} + {\left(y - 2\right)}^{2} / 16 = \frac{16}{16}$

${\left(x - 0\right)}^{2} / 4 + {\left(y - 2\right)}^{2} / 16 = 1$

Kindly see the graph

graph{(x-0)^2/4+(y-2)^2/16=1[-20,20,-10,10]}

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