Find the condition that

$\displaystyle ax^2 + 2hxy + by^2$

is resolvable into linear factors of the form $\displaystyle (y - mx)$ and $\displaystyle (my + x)$.

Not a clue on this one....

2. \displaystyle \displaystyle \begin{align*}(y - mx)(my + x) &= my^2 + xy - m^2xy - mx^2 \\ &= -mx^2 + (1 - m^2)xy + my^2\end{align*}

If this is equal to $\displaystyle \displaystyle ax^2 + 2hxy + by^2$ then $\displaystyle \displaystyle a = -m, 2h = 1 - m^2, b = m$.

3. But the question says that there must be be only ONE single condition , not three . But from the three given by you...it can be converted into two...which are.."Since $\displaystyle b = m$ and $\displaystyle a = - m$ , $\displaystyle a + b = 0$." and $\displaystyle 2h = 1 - m^2$ . The answer given at the back of the book is $\displaystyle a + b = 0$ . But the $\displaystyle 2h = 1 - m^2$ can't simply be ignored , can it ?

4. I've got it !

If $\displaystyle (y - mx)$ be a factor of $\displaystyle ax^2 + 2hxy + by^2$ then it must vanish for $\displaystyle y = mx$.

Putting $\displaystyle y = mx$ in $\displaystyle ax^2 + 2hxy + by^2$ , we get $\displaystyle ax^2 + 2hmx^2 + b(m^2)(x^2) = 0$ , or $\displaystyle a + 2hm + bm^2 = 0$ .

Similarly if $\displaystyle ( my + x)$ be a factor of $\displaystyle ax^2 + 2hxy + by^2$ then it must vanish for $\displaystyle x = - my$.

Putting $\displaystyle x = - my$ in $\displaystyle ax^2 + 2hxy + by^2$ , we get $\displaystyle a(m^2)(y^2) - 2hmy^2 + by^2 = 0$ or $\displaystyle am^2 - 2hm + b = 0$ .

Then since $\displaystyle m$ satisfies both the equations

$\displaystyle bm^2 + 2hm + a = 0$.....................$\displaystyle (1)$
and, $\displaystyle am^2 - 2hm + b = 0$.......................$\displaystyle (2)$

The condition which is required for both the above equations to have both the roots common is actually the required condition .

On taking some determinants we easily get that, the required condition is

$\displaystyle \frac{a}{b } = \frac{-2h}{2h} = \frac{b}{a}$
$\displaystyle \Rightarrow \frac{a}{b } = -1 = \frac{b}{a}$
$\displaystyle \Rightarrow a + b = 0$ Q.E.D