If the pair of lines x^2-16pxy-y^2=0 and x^2- | vedclass

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(A) Let the given equations be:$x^2-16pxy-y^2=0 \quad (i)$$x^2-16qxy-y^2=0 \quad (ii)$The equation of the angle bisectors of a pair of lines $ax^2+2hx

Vedclass · Accepted Answer

$\frac{-1}{64}$

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(A) Let the given equations be:$x^2-16pxy-y^2=0 \quad (i)$$x^2-16qxy-y^2=0 \quad (ii)$The equation of the angle bisectors of a pair of lines $ax^2+2hxy+by^2=0$ is given by $\frac{x^2-y^2}{a-b} = \frac{xy}{h}$.For equation $(i)$,$a=1, b=-1, h=-8p$. The bisectors are $\frac{x^2-y^2}{1-(-1)} = \frac{xy}{-8p} \implies \frac{x^2-y^2}{2} = \frac{xy}{-8p} \implies -8px^2+8py^2=2xy \implies -8px^2-2xy+8py^2=0 \quad (iii)$.For equation $(ii)$,$a=1, b=-1, h=-8q$. The bisectors are $\frac{x^2-y^2}{1-(-1)} = \frac{xy}{-8q} \implies \frac{x^2-y^2}{2} = \frac{xy}{-8q} \implies -8qx^2-2xy+8qy^2=0 \quad (iv)$.Since each pair bisects the angle between the other,equation $(i)$ must be the same as equation $(iv)$ and equation $(ii)$ must be the same as equation $(iii)$.Comparing $(i)$ and $(iv)$: $\frac{1}{-8q} = \frac{-16p}{-2} = \frac{-1}{8q}$. This leads to $1 = -64pq$,so $pq = \frac{-1}{64}$.

If the pair of lines $x^2-16pxy-y^2=0$ and $x^2-16qxy-y^2=0$ are such that each pair bisects the angle between the other pair,then $pq=$

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