Sum and Product of Slopes of a Pair

For the pair ax²+2hxy+by²=0, the two slopes satisfy these Vieta-like relations.

For the pair $ax^2 + 2hxy + by^2 = 0$, divide by $x^2$ (assuming $x \neq 0$):

$$b\left(\frac{y}{x}\right)^2 + 2h\left(\frac{y}{x}\right) + a = 0$$

The slopes $m_1 = y/x$ and $m_2 = y/x$ of the two lines are the roots of:

$$bm^2 + 2hm + a = 0$$

By Vieta's formulas:

$$\boxed{m_1 + m_2 = -\frac{2h}{b}, \qquad m_1 m_2 = \frac{a}{b}}$$

Why this is useful

Many JEE problems give a condition on the slopes (e.g., one slope is twice the other, slopes are in ratio $1:2$, sum of squares of slopes equals a constant) and ask for a relation between $a$, $h$, $b$. These reduce to one or two equations using $m_1 + m_2$ and $m_1m_2$ — without ever finding $m_1$ and $m_2$ individually.

Example: If one slope is double the other, $m_2 = 2m_1$:

$$m_1 + 2m_1 = -\frac{2h}{b} \Rightarrow m_1 = -\frac{2h}{3b}$$ $$m_1 \cdot 2m_1 = \frac{a}{b} \Rightarrow 2m_1^2 = \frac{a}{b}$$

Substituting $m_1$: $2 \cdot \frac{4h^2}{9b^2} = \frac{a}{b}$, giving $8h^2 = 9ab$.