If d is the distance between the point of intersection of the lines x² + 4xy + ky² - 4x - 10y + 3 = 0 and the origin and p is the product of the perpendicular distances from the origin to these lines, then d² - 20p² =

x² + 4xy + ky - 4x -10y + 3 = 0 represents a pair of straight lines
(1)(k)(3) + 2(2)(-2)(-5)-1(-5)² -k(-2)² -3(2)²=0
3k + 40 - 25 - 4k -12 = 0→k = 3
Point of intersection $=\left(\frac{\mathrm{hf}-\mathrm{bg}}{\mathrm{ab}-{\mathrm{h}}^2},\frac{\mathrm{gh}-\mathrm{af}}{\mathrm{ab}-{\mathrm{h}}^2}\right)=\left(\frac{2(-5)-3(-2)}{3-4},\frac{(-2)\left(2\right)-(-5)}{3-4}\right)=(4-1)$
$\begin{array}{l}\therefore \mathrm{d}=\sqrt{(4-0{)}^2+(-1-0{)}^2}=\sqrt{16+1}=\sqrt{17}\\ \mathrm{p}=\frac{3}{\sqrt{(1-3{)}^2+(4{)}^2}}=\frac{3}{\sqrt{20}}\\ \therefore {\mathrm{d}}^2-20{\mathrm{p}}^2=17-20\left(\frac{9}{20}\right)=17-9=8\end{array}$