Equation $a x^{3} - 9 y x^{2} - y^{2} x + 4 y^{3} = 0$ represents three straight lines. If two of the lines are perpendicular, then $a = . . . . .$

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Solution

The given equation being homogeneous equation of third
degree represents three straight lines through the origin.
Since two of them are perpendicular, they are given by
$x^{2} + p x y - y^{2} = 0$
Hence the given equation is
$(a x - 4 y) (x^{2} + p x y - y^{2}) = 0$
Comparing coefficient, we get
$a p - 4 = - 9, - a - 4 p = - 1$ .
$∴ p = - \frac{5}{a} = \frac{a - 1}{- 4}$
or $a^{2} - a - 20 = 0 \Rightarrow (a - 5) (a + 4) = 0$
$∴ a = 5, - 4$

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