First slide

Pair of straight lines

Question

If the lines represented by the equation $3 y^{2} - x^{2} + 2 \sqrt{3} x - 3 = 0$ are rotated about the point $(\sqrt{3}, 0)$ through an angle of $15^{0},$ one in clockwise direction and the other in anticlockwise direction, so that they become perpendicular, then the equation of the pair of lines in the new position is

Moderate

A

$y^{2} - x^{2} + 2 \sqrt{3} x + 3 = 0$
B

$y^{2} - x^{2} + 2 \sqrt{3} x - 3 = 0$
C

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

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

Solution

The given equation of the pair of straight lines can be rewritten as $(\sqrt{3} y - x + \sqrt{3}) (\sqrt{3} y + x - \sqrt{3}) = 0$
Their separate equations are $\sqrt{3} y - x + \sqrt{3} = 0$ and $\sqrt{3} y + x - \sqrt{3} = 0$
or $y = \frac{1}{\sqrt{3}} x - 1$ and $y = - \frac{1}{\sqrt{3}} x + - 1$
or $y = (\tan 30^{0}) x - 1$ and $y = (\tan 150^{0}) x + 1$
After rotation through an angle of $15^{0}$ as given in the question, the lines are $(y - 0) = \tan 45^{0} (x - \sqrt{3})$ and $(y - 0) = \tan 135^{0} (x - \sqrt{3})$
or $y = x - \sqrt{3}$ $y = - x + \sqrt{3}$ Their combined equation is $(y - x + \sqrt{3}) (y + x - \sqrt{3}) = 0$
or $y^{2} - x^{2} + 2 \sqrt{3} x - 3 = 0$

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