Explanation:
(D) : Given the equation,
\(4 x^{2}-y^{2}-8 x+4 y=0\)
Comparing the equation, \(a x^{2}+2 h x y+b y^{2}+2 g x+2 f y\) \(+\mathrm{c}=0\)
Then, \(\mathrm{a}=2, \mathrm{~b}=-1\)
\(\mathrm{g}=-4, \mathrm{f}=2, \mathrm{~h}=0, \mathrm{c}=0\)
Point of intersection \(=\)
\(\left(\frac{\mathrm{hf}-\mathrm{bg}}{\mathrm{ab}-\mathrm{h}^{2}}, \frac{\mathrm{gh}-\mathrm{af}}{\mathrm{ab}-\mathrm{h}^{2}}\right)=\left(\frac{0-4}{-2}, \frac{0-4}{-2}\right) \equiv(2,2)\)