First slide

Pair of straight lines

Question

If the pairs of lines $x^{2} + 2 xy + {ay}^{2} = 0$ and ${ax}^{2} + 2 xy + y^{2} = 0$ have exactly one line in common, then the join t equation of the other two lines is given by

Moderate

A

$3 x^{2} + 8 xy - 3 y^{2} = 0$
B

$3 x^{2} + 10 xy + 3 y^{2} = 0$
C

$y^{2} + 2 xy - 3 x^{2} = 0$
D

$x^{2} + 2 xy - 3 y^{2} = 0$

Solution

Let $y = mx$ be a line common to the given pairs of lines.
Then, ${am}^{2} + 2 m + 1 = 0$ and $m^{2} + 2 m + a = 0$
$∴ \frac{m^{2}}{2 (1 - a)} = \frac{m}{a^{2} - 1} = \frac{1}{2 (1 - a)}$
or $m^{2} = 1 and m = - \frac{a + 1}{2}$
or ${(a + 1)}^{2} = 4$
or a=1
or -3
But for a=1, the two pairs have both the lines common.
So $a = - 3$ and the slope m of the line common to both the pairs is 1.
Now, $∴ x^{2} + 2 xy + {ay}^{2} = x^{2} + 2 xy - 3 y^{2} = (x - y) (x + 3 y)$ and ${ax}^{2} + 2 xy + y^{2} = - 3 x^{2} + 2 xy + y^{2} = - (x - y) (3 x + y)$
So, the equation of the required pair of lines is $(x + 3 y) (3 x + y) = 0$
or $3 x^{2} + 10 xy + 3 y^{2} = 0$

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