First slide

Pair of straight lines

Question

If the slope of one of the lines represented by $a x^{2} + 2 h x y + b y^{2} = 0$ be the square of the other, then $\frac{a + b}{h} + \frac{8 h^{2}}{a b}$ is equal to

Moderate

A

0
B

1
C

6
D

8

Solution

We can write ${(\frac{y}{x})}^{2} + \frac{2 h}{b} (\frac{y}{x}) + \frac{a}{b} = 0$
If the slopes of the lines are $m$ and $m^{2}$ then
$\begin{array}{l} m + m^{2} = - \frac{2 h}{b} and m m^{2} = \frac{a}{b} \\ \Rightarrow {(\frac{a}{b})}^{1 / 3} + {(\frac{a}{b})}^{2 / 3} = - \frac{2 h}{b} \\ \Rightarrow \frac{a}{b} + {(\frac{a}{b})}^{2} + 3 (\frac{a}{b}) (- \frac{2 h}{b}) = - \frac{8 h^{3}}{b^{3}} \\ \Rightarrow \frac{a}{b^{2}} [b + a - 6 h] + \frac{8 h^{3}}{b^{3}} = 0 \end{array}$
$\begin{array}{l} \Rightarrow a + b - 6 h + \frac{8 h^{3}}{a b} = 0 \\ \Rightarrow \frac{a + b}{h} + \frac{8 h^{2}}{a b} = 6 . \end{array}$

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