Question

What does the conic:
x^2+12xy-4y^2-6x+4y+9=0

Answer 1

Answer:

Given conic is x2−6x+4y+1=0

Given conic is x2−6x+4y+1=0⟹x2−6x+9−9+4y+1=0

Given conic is x2−6x+4y+1=0⟹x2−6x+9−9+4y+1=0⟹x2−3x−3x+9−9+4y+1=0

Given conic is x2−6x+4y+1=0⟹x2−6x+9−9+4y+1=0⟹x2−3x−3x+9−9+4y+1=0⟹x(x−3)−3(x−3)+4y−8=0

Given conic is x2−6x+4y+1=0⟹x2−6x+9−9+4y+1=0⟹x2−3x−3x+9−9+4y+1=0⟹x(x−3)−3(x−3)+4y−8=0⟹(x−3)2+4y−8=0

Given conic is x2−6x+4y+1=0⟹x2−6x+9−9+4y+1=0⟹x2−3x−3x+9−9+4y+1=0⟹x(x−3)−3(x−3)+4y−8=0⟹(x−3)2+4y−8=0⟹(x−3)2=−4y+8

Given conic is x2−6x+4y+1=0⟹x2−6x+9−9+4y+1=0⟹x2−3x−3x+9−9+4y+1=0⟹x(x−3)−3(x−3)+4y−8=0⟹(x−3)2+4y−8=0⟹(x−3)2=−4y+8⟹(x−3)2=−4(y−2)

Given conic is x2−6x+4y+1=0⟹x2−6x+9−9+4y+1=0⟹x2−3x−3x+9−9+4y+1=0⟹x(x−3)−3(x−3)+4y−8=0⟹(x−3)2+4y−8=0⟹(x−3)2=−4y+8⟹(x−3)2=−4(y−2)⟹(x−3)2=4(−1)(y−2)

Given conic is x2−6x+4y+1=0⟹x2−6x+9−9+4y+1=0⟹x2−3x−3x+9−9+4y+1=0⟹x(x−3)−3(x−3)+4y−8=0⟹(x−3)2+4y−8=0⟹(x−3)2=−4y+8⟹(x−3)2=−4(y−2)⟹(x−3)2=4(−1)(y−2)Thus, the focus is (3,1).