Suppose the pairs of straight lines x^2 - 2ax | vedclass

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(B) The equation of a pair of lines is $Ax^2 + 2Hxy + By^2 = 0$. The angle bisectors of this pair are given by $\frac{x^2 - y^2}{A - B} = \frac{xy}{H}

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$-1$

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(B) The equation of a pair of lines is $Ax^2 + 2Hxy + By^2 = 0$. The angle bisectors of this pair are given by $\frac{x^2 - y^2}{A - B} = \frac{xy}{H}$.For the first pair $x^2 - 2axy - y^2 = 0$,we have $A=1, B=-1, H=-a$. The bisectors are $\frac{x^2 - y^2}{1 - (-1)} = \frac{xy}{-a}$,which simplifies to $\frac{x^2 - y^2}{2} = \frac{xy}{-a}$,or $ax^2 + 2xy - ay^2 = 0$.Since this must be the second pair $x^2 - 2bxy - y^2 = 0$,we compare the coefficients:$\frac{a}{1} = \frac{2}{-2b} = \frac{-a}{-1}$.From $\frac{a}{1} = \frac{-a}{-1}$,we get $a=a$ (always true).From $\frac{a}{1} = \frac{2}{-2b}$,we get $a = -\frac{1}{b}$,which implies $ab = -1$.

Suppose the pairs of straight lines $x^2 - 2axy - y^2 = 0$ and $x^2 - 2bxy - y^2 = 0$ are such that each pair bisects the angles between the other. Then $ab =$

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