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.