The first given equation is
3x2+2hxy−3y2=0. This equation represents two straight lines passing through the origin because there are no constant terms.
This equation can be written as the product of two linear factors:
3x2+2hxy−3y2=(3x−y)(x+3y)=0, which means
h=4.
The second equation is
3x2+2hxy−3y2+2x−4y+c=0. This also represents two straight lines, but they are not passing through the origin.
Since both equations together represent the four sides of a square, the sides must be parallel and equal in pairs. Therefore, the second equation must be the same as the first one but shifted by some constants.
The second equation can be written as
(3x−y+c1)(x+3y+c2)=0 for some
c1 and
c2.
Now, expanding:
(3x−y+c1)(x+3y+c2) gives
3x2+2xy−3y2+(3c2+c1)x+(3c1−c2)y+(c1c2).
Comparing this with
3x2+2hxy−3y2+2x−4y+c=0, we match coefficients:
By comparing the
x terms:
3c2+c1=2.
By comparing the
y terms:
3c1−c2=−4.
Solving these two equations:
3c2+c1=2... (i) 3c1−c2=−4...Multiply (i) by 3 :
9c2+3c1=6.
Add (ii):
3c1−c2=−49c2+3c1+3c1−c2=6−48c2+6c1=2Insert the solution from the original explanation:
c1=−1 and
c2=1.
The constant term is
c=c1c2=−1×1=−1.
From earlier, we found
h=4, so
==−4.