if pair of lines ax^2+2hxy+by^2+2gx+2fy+c=0 intersect on y axis then ... a.2fgh=hg^2+ch^2b.hg^2 not equal to ch^2c.abc=2fghd.none of these

A pair of equations intersecting on the y-axis is given by:
(px+my+n)(qx+my+n)=0. You can see that by putting x=0 in each factor on the LHS, and observing that the y-coordinate is the same. Expanding and equating the coefficients of x^2, xy, etc in the given expression, we have
a=pq, 2h=m(p+q), b=m^2, 2g=n(p+q), f=mn and c=n^2.
Now the reference cited below proves the general condition for a second-degree equation to represent a pair of straight lines as:
af^2 + bg^2 + ch^2 = 2fgh + abc (You can go through the proof if you wish)
Using our results, af^2 =pqm^2n^2 = abc. Thus af^2 and abc cancel out from the general condition, and we're left with bg^2 + ch^2 = 2fgh, which is what we're asked to prove.
Thanks
Bharat Bajaj
IIT Delhi