If the slopes of the lines represented by Kx^ | vedclass

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(A) The equation of the pair of lines is $Kx^2 - 4xy + 5y^2 = 0$. Comparing this with $ax^2 + 2hxy + by^2 = 0$,we get $a = K$,$2h = -4 \Rightarrow h =

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$\frac{-21}{5}$

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(A) The equation of the pair of lines is $Kx^2 - 4xy + 5y^2 = 0$. Comparing this with $ax^2 + 2hxy + by^2 = 0$,we get $a = K$,$2h = -4 \Rightarrow h = -2$,and $b = 5$. Let $m_1$ and $m_2$ be the slopes of the lines. The difference between the slopes is given by $|m_1 - m_2| = \frac{2\sqrt{h^2 - ab}}{|b|}$. Given $|m_1 - m_2| = 2$,we have: $2 = \frac{2\sqrt{(-2)^2 - K(5)}}{5}$ $1 = \frac{\sqrt{4 - 5K}}{5}$ $5 = \sqrt{4 - 5K}$ Squaring both sides: $25 = 4 - 5K$ $5K = 4 - 25$ $5K = -21$ $K = \frac{-21}{5}$

If the slopes of the lines represented by $Kx^2 - 4xy + 5y^2 = 0$ differ by $2$,then $K = $

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