Question

y=mx is one of bisect of linex^2-2y^2-8xy=0 the value of m​

Answer 1

Pair of Straight Lines

Formula: If $ax^{2}+2hxy+by^{2}=0$ be a pair of straight lines, then the equation of the pair of its bisectors is given by

$\quad \frac{x^{2}-y^{2}}{a-b}=\frac{xy}{h}$.

Solution:

The given pair of straight lines is

$\quad x^{2}-8xy-2y^{2}=0$

Comparing it with the above equation of pair of straight lines, we get

$\quad a=1,\:h=-4,\:b=-2$

So the pair of its bisectors is

$\quad \frac{x^{2}-y^{2}}{1-(-2)}=\frac{xy}{-4}$

$\Rightarrow \frac{x^{2}-y^{2}}{3}=-\frac{xy}{4}$

$\Rightarrow 4x^{2}-4y^{2}+3xy=0$

$\Rightarrow 4\frac{x^{2}}{x^{2}}-4\frac{y^{2}}{x^{2}}+3\frac{xy}{x^{2}}=0$

$\Rightarrow 4-4m^{2}+3m=0\quad[\because y=mx\:]$

$\Rightarrow 4m^{2}-3m-4=0$

Using quadratic formula, we get

$\quad m=\frac{-(-3)\pm\sqrt{(-3)^{2}-4 (4)(-4)}}{2times 4}$

$\Rightarrow m=\frac{3\pm\sqrt{73}}{8}$

This is the required value of $m$.

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