Question

Find the conditions that the expressions
$a{x}^2+2hxy+b{y}^2$, $a^{\prime }{x}^2+2h^{\prime }xy+b^{\prime }{y}^2$ may be respectively divisible by factors of the form $y-mx,my+x.$

Answer 1

Multiple the given two expressions and then substitute $y=mx$ and $my=-x$ to get $m^{4}a{a}^{\prime }+m^3(2a{h}^{\prime }+2a^{\prime }h)+m^2(a^{\prime }b+b^{\prime }a+4h{h}^{\prime })+m(2h^{\prime }b+2b^{\prime }h)+b{b}^{\prime }=0$

And $a{a}^{\prime }-m(2a{h}^{\prime }+2a^{\prime }h)+m^2(a^{\prime }b+b^{\prime }a+4h{h}^{\prime })-m^3(2h^{\prime }b+2b^{\prime }h)+b{b}^{\prime }{m}^4=0$

Now subtracting both the equations, we get

$m^4(a{a}^{\prime }-b{b}^{\prime })+m^3(2a{h}^{\prime }+2a^{\prime }h+2b{h}^{\prime }+2b^{\prime }h)+m(2h^{\prime }b+2b^{\prime }h+2h^{\prime }a+2a^{\prime }h)+b{b}^{\prime }-a{a}^{\prime }=0$

Taking out common factor and simplifying yeils $(m^2+1)\left(\right(a{a}^{\prime }-b{b}^{\prime }\left)\right(m^2-1)+2m(h{a}^{\prime }+h^{\prime }a+h{b}^{\prime }+h^{\prime }b\left)\right)=0$

And since this equation must be satisfied by a single $m$, so determinant must be $0$ giving us $4h^2(a+a^{\prime }+b+b^{\prime }{)}^2-4(a{a}^{\prime }-b{b}^{\prime }{)}^2=0$.