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The angle between the lines represented by the equation $ax^2+2hxy-by^2+2gx+2fy+c=0$ is:

$1$. $90^\circ $

$2$. $60^\circ $

$3$. $\tan^{-1} (\dfrac {2\sqrt {h^2-ab}}{ab}$

$4$. $30^\circ $.

My Attempt: $$ax^2+2hxy-by^2+2gx+2fy+c=0$$ Taking homogeneous equation: $$ax^2+2hxy-by^2=0$$ Let $\theta $ be the angle between the pair of lines. Then, $$\tan \theta =\pm \dfrac {2\sqrt {h^2-ab}}{a+b}$$.

How do I proceed further?

pi-π
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  • Since the equation of the intersecting lines is completely generic, the only possible choice is 3. The solution you came up with differs from that only in the denominator, so either there’s a typo in that answer or you made some error in your computation. – amd May 09 '17 at 02:31

1 Answers1

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The answer will be $$\tan \theta =\pm \frac {2\sqrt {h^2-ab}}{a+b}$$

This is so, since the angle between doesn't depend on shifting of origin, we can shift origin to the point of intersection of the lines.

Let the point of intersection of the lines given by : $$ax^2+2hxy-by^2+2gx+2fy+c=0$$

Be $(\alpha,\beta)$ now on moving the origin to $(\alpha,\beta)$, replace $x \to x-\alpha$ and $y \to y-\alpha$, doing so we get the transformation of old equation into new homogeneous equation, but, since the angle between pair of straight lines represented by homogeneous equation depends only on the coefficient of $x^2,y^2$ and $xy$, the angle will remain same.

Jaideep Khare
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