Question
If the straight lines joining the origin and the points of intersection of the curve $5x^{2} + 12xy - 6y^{2} + 4x - 2y + 3 = 0$ and $x + ky - 1 = 0$ are equally inclined to the $x$-axis then the value of $k$ is :
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Solution
Sol. [Key Concept: Homogenize the conic with the line to get the pair of lines through the origin; equal inclination to the $x$-axis ⇒ coefficient of $xy = 0$.]
Equation of pair $OP$ and $OQ$ is obtained by homogenising.
⇒ Equation of pair $OP$ and $OQ$ is
$$\begin{matrix} 5x^{2} + 12xy - 6y^{2} + (4x - 2y)(x + ky) \\ + 3(x + ky)^{2} = 0 \end{matrix}$$
$OP$ and $OQ$ are equally inclined to $x$-axis.
Coefficient of $xy = 0$
$$\begin{matrix} \Rightarrow & 12 + 4k - 2 + 6k & = 0 \\ \Rightarrow & & k = - 1 \end{matrix}$$
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