Question

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

Answer 1

Vertical ellipse
Equation: `(x-2)^2/(4)+ (y-3)^2/(16) = 1`

Explanation:

We can find the equation by completing the square

`4x^2 +y^2 -16x-6y+9 = 0`

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

`4(x^2-4x +color(red)square) + (y^2 -6y +color(blue)square) = -9 +4color(red)(*(square)) + color(blue)(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(x-2)^2+ (y-3)^2 = 16`

`(4(x-2)^2)/(16)+ (y-3)^2/(16) = 16/16`

`(x-2)^2/(4)+ (y-3)^2/(16) = 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]}