Question

Find the value of k, if the lines are represented by $k({{x}^{2}}+{{y}^{2}})=8xy$ are coincident.

Answer 1

Hint: The given problem is related to pairs of straight lines. For the homogeneous equation of pair of straight lines $a{{x}^{2}}+b{{y}^{2}}+2hxy=0~$, condition of coincidence is ${{h}^{2}}-ab=0$ . Compare the given equation of lines with $a{{x}^{2}}+b{{y}^{2}}+2hxy=0~$to get the values of a, b and h. Then, substitute these values in ${{h}^{2}}-ab=0$ to get the value of k.

Complete step-by-step answer:

The equation of lines given to us is $k({{x}^{2}}+{{y}^{2}})=8xy$ , which can also be written as $k{{x}^{2}}+k{{y}^{2}}=8xy$.

Now the given equation is, $k{{x}^{2}}+k{{y}^{2}}=8xy$.
We know, the homogeneous equation of a pair of straight lines is given as $a{{x}^{2}}+b{{y}^{2}}+2hxy=0~$. Now, we will compare the equation with the $a{{x}^{2}}+b{{y}^{2}}+2hxy=0~$.
Form the comparison of the ${{x}^{2}}$coefficient:
$\Rightarrow a=k$
Form the comparison of the ${{y}^{2}}$coefficient:
$\Rightarrow b=k$
Form the comparison of the $xy$coefficient:
$\Rightarrow h=-4$
Now, we know, for the homogeneous equation of pair of straight lines $a{{x}^{2}}+b{{y}^{2}}+2hxy=0~$, condition of coincidence is ${{h}^{2}}-ab=0$ . We will substitute the values of a, b and h in ${{h}^{2}}-ab=0$, to get the value of the k.
${{\left( -4 \right)}^{2}}-{{k}^{2}}=0$
$\Rightarrow {{k}^{2}}=16$
$\Rightarrow k=\pm 4$
Therefore, the values of k are 4, -4.

Note: An equation of the type $a{{x}^{2}}~+\text{ }2hxy\text{ }+\text{ }b{{y}^{2}}~$ which is a homogeneous equation of degree 2 denotes a pair of straight lines passing through the origin. The lines may be real, coincident or imaginary depending on the conditions satisfied by them. The condition are:
i.If ${{h}^{2}}~>\text{ }ab$, then the lines are real and distinct.
ii.If ${{h}^{2}}~<\text{ }ab$, then the lines are imaginary with the point of intersection as (0, 0).
iii.If ${{h}^{2}}~=\text{ }ab$, then the lines are coincident.