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

1 Answer
Nov 29, 2015

Answer:

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]}