Prove that x^2+6xy+9y^2+4x+12y-5=0 represents a pair of parallel straight lines and find the distance between them.

Question

amistre64 · Answer

might want to split it into x and y parts on each side and complete some squares?  just an idea

amistre64 · Answer

or, use the the x^2,xy,y^2 parts to complete a square with to compare it to the linear aspects

anonymous · Answer

sory i didnt get :O

amistre64 · Answer

$$x^2+6xy+9y^2+4x+12y-5=0$$
$$(x^2+6xy+9y^2)=-4x-12y+5$$
$$(x^2+2(3xy)+(3y)^2)=-4x-12y+5$$
$$(x+3y)^2=-4x-12y+5$$
yeah, im not sure how a conic section will construct 2 parallel lines either ...

amistre64 · Answer

http://www.wolframalpha.com/input/?i=x%5E2%2B6xy%2B9y%5E2%2B4x%2B12y-5%3D0 but it is ....

amistre64 · Answer

y = mx,  y= a(mx+b)

mx = a(mx+b)

m^2x^2 = a^2(m^2x^2+b^2+2bmx

m^2x^2 = a^2m^2x^2+a^2b^2+2ba^2mx

m^2x^2 - a^2m^2x^2 - 2ba^2mx - a^2b^2 = 0

y^2 - (a^2m^2)x^2 - 2(ba^2m)x - a^2b^2 = 0

.....