The equation ${{x}^{2}}+k{{y}^{2}}+4xy=0$ represents two coincident lines, if $k=$
A. $0$
B. $1$
C. $4$
D. $16$

Hint: In this question, we are to find the value of the variable $k$ in the given pair of straight lines. Since it is given that the two lines are coincident lines, we can write ${{h}^{2}}=ab$. By applying this, we can find the required value from the given equation.



Formula Used:The combined equation of pair of straight lines is written as
$H\equiv a{{x}^{2}}+2hxy+b{{y}^{2}}=0$
This is called a homogenous equation of the second degree in $x$ and $y$
And
$S\equiv a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0$
This is called a general equation of the second degree in $x$ and $y$.
If ${{h}^{2}}If ${{h}^{2}}=ab$, then $a{{x}^{2}}+2hxy+b{{y}^{2}}=0$ represents coincident lines.
If ${{h}^{2}}>ab$, then $a{{x}^{2}}+2hxy+b{{y}^{2}}=0$ represents two real and different lines that pass through the origin.
Thus, the equation $a{{x}^{2}}+2hxy+b{{y}^{2}}=0$ represents two lines. They are:
$ax+hy\pm y\sqrt{{{h}^{2}}-ab}=0$



Complete step by step solution:Given equation is
${{x}^{2}}+k{{y}^{2}}+4xy=0$
We know that a pair of straight lines given by the equation $a{{x}^{2}}+2hxy+b{{y}^{2}}=0$ are coincident lines if they satisfy ${{h}^{2}}=ab$.
So, by comparing the given equation and the general equation of the pair of straight lines, we get
$a=1;b=k;h=2$
Then, substituting these values in the above condition, we get
\[\begin{align}
  & {{h}^{2}}=ab \\
 & \Rightarrow {{2}^{2}}=(1)(k) \\
 & \therefore k=4 \\
\end{align}\]
So, the given equation we can rewrite as ${{x}^{2}}+4{{y}^{2}}+4xy=0$.



Option ‘C’ is correct


Note: Here, we need to remember that the given lines are coincident and given in the homogenous form as $a{{x}^{2}}+2hxy+b{{y}^{2}}=0$. So, we can apply the condition ${{h}^{2}}=ab$ for these coincident lines in order to find the required variable and its value.