1. Tardigrade
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  4. If the straight lines joining the origin and the points of intersection of the curve 5 x2+12 x y-6 y2+4 x-2 y+3=0 and x+k y-1=0 are equally inclined to the co-ordinate axis, then the value of k -

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Q. If the straight lines joining the origin and the points of intersection of the curve $5 x^2+12 x y-6 y^2+4 x-2 y+3=0$ and $x+k y-1=0$ are equally inclined to the co-ordinate axis, then the value of $k$ -

Straight Lines

Solution:

Homogenizing the curve with the help of the straight line.
$5 x^2+12 x y-6 y^2+4 x(x+k y)-2 y(x+k y)+3(x+k y)^2=0 $
$12 x^2+(10+4 k+6 k) x y+\left(3 k^2-2 k-6\right) y^2=0$
Lines are equally inclined to the coordinate axes
$\therefore \text { coefficient of } xy =0$
$\Rightarrow 10 k +10=0 \Rightarrow k =-1$