Find the value of $k$ , if slope of one of the lines given by $k x^{2} + 4 x y - y^{2} = 0$ exceeds the slope of the other by $8$ .

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Solution

According to the problem:-
consider the given joint equation $= k x^{2} + 4 x y - y^{2}$

comparing with $a x^{2} + 2 h x y + b y^{2} = 0$
We get,
$a = k, 2 h = 4, b = - 1$

Let the slopes of the two lines represented by given joint equation is $m_{1} a n d m_{2}$

$\Rightarrow$ $(m_{1} + m_{2}) = \frac{- 2 h}{b}$ $= \frac{- 4}{- 1}$
$\Rightarrow$ $m_{1} + m_{2} = 4$

and
$m_{1} . m_{2} = d \frac{a}{b}$ $= \frac{k}{- 1}$ $= - k$

Given that ,
$| m_{1} - m_{2} |$ =8

squaring both side
${(m_{1} - m_{2})}^{2} = 64$

since ${(m_{1} + m_{2})}^{2} - 4 m_{1} \cdot m_{2} = 64$

$\Rightarrow$ $4^{2} - 4 (- k) = 64$

$\Rightarrow$ $4 k = 48$

$\Rightarrow$ $k = \frac{48}{4}$

$\Rightarrow$ $k = 12$

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