# How do you identify the conic section represented by each equation 4x^2+y^2-16x-6y+9=0?

Nov 29, 2015

Vertical ellipse
Equation: ${\left(x - 2\right)}^{2} / \left(4\right) + {\left(y - 3\right)}^{2} / \left(16\right) = 1$

#### Explanation:

We can find the equation by completing the square

$4 {x}^{2} + {y}^{2} - 16 x - 6 y + 9 = 0$

$\left(4 {x}^{2} - 16 x + \textcolor{red}{\square}\right) + \left({y}^{2} - 6 y + \textcolor{b l u e}{\square}\right) = - 9 + \textcolor{red}{\square} + \textcolor{b l u e}{\square}$

$4 \left({x}^{2} - 4 x + \textcolor{red}{\square}\right) + \left({y}^{2} - 6 y + \textcolor{b l u e}{\square}\right) = - 9 + 4 \textcolor{red}{\cdot \left(\square\right)} + \textcolor{b l u e}{\square}$

4(x^2-4x +color(red)4)+ (y^2 -6y color(blue)(+9) )= -9 +4color(red)(*(4) + color(blue)(9)

4(x^2-4x +color(red)4)+ (y^2 -6y color(blue)(+9)) = -9 +color(red)(*(16) + color(blue)(9)

$4 {\left(x - 2\right)}^{2} + {\left(y - 3\right)}^{2} = 16$

$\frac{4 {\left(x - 2\right)}^{2}}{16} + {\left(y - 3\right)}^{2} / \left(16\right) = \frac{16}{16}$

${\left(x - 2\right)}^{2} / \left(4\right) + {\left(y - 3\right)}^{2} / \left(16\right) = 1$

This is a vertical ellipse with center at #(2, 3)

graph{4x^2+y^2-16x-6y+9=0 [-8.59, 9.19, -1.186, 7.704]}