Question

If in general quadratic equation $f(x,y)=0$, $\Delta =0$ and ${{h}^{2}}=ab$, then the equation represents
A. Two parallel straight lines
B. Two perpendicular straight lines
C. Two coincident lines
D. None of these

Answer 1

Hint: In this question, we are to find the type of the given quadratic function that represents a pair of lines. For this, the given conditions are applied in the quadratic function. So, on substituting them we get the type of those lines.



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:
$ax+hy\pm y\sqrt{{{h}^{2}}-ab}=0$
If $S\equiv a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0$ represents a pair of lines, then
i) $abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}=0$ and
ii) ${{h}^{2}}\ge ab,{{g}^{2}}\ge ac,{{f}^{2}}\ge bc$



Complete step by step solution:Given quadratic function is,
$f(x,y)=0$
That is,
$f(x,y)=S=a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0$
This equation represents pair of lines if $\Delta =0$.
Here, $\Delta =abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}=0$
Consider the two lines as
\[\begin{align}
  & {{l}_{1}}x+{{m}_{1}}y+{{n}_{1}}=0\text{ }...(1) \\
 & {{l}_{2}}x+{{m}_{2}}y+{{n}_{2}}=0\text{ }...(2) \\
\end{align}\]
For the combined equation, multiplying these two equations, we get
\[\begin{align}
  & ({{l}_{1}}x+{{m}_{1}}y+{{n}_{1}})({{l}_{2}}x+{{m}_{2}}y+{{n}_{2}}) \\
 & \Rightarrow {{l}_{1}}{{l}_{2}}{{x}^{2}}+{{l}_{1}}{{m}_{2}}xy+{{l}_{1}}{{n}_{2}}x+{{l}_{2}}{{m}_{1}}xy+{{m}_{1}}{{m}_{2}}{{y}^{2}}+{{m}_{1}}{{n}_{2}}y+{{l}_{2}}{{n}_{1}}x+{{n}_{1}}{{m}_{2}}y+{{n}_{1}}{{n}_{2}} \\
 & \Rightarrow {{l}_{1}}{{l}_{2}}{{x}^{2}}+({{l}_{1}}{{m}_{2}}+{{l}_{2}}{{m}_{1}})xy+{{m}_{1}}{{m}_{2}}{{y}^{2}}+({{l}_{1}}{{n}_{2}}+{{l}_{2}}{{n}_{1}})x+({{m}_{1}}{{n}_{2}}+{{n}_{1}}{{m}_{2}})y+{{n}_{1}}{{n}_{2}}\text{ }...(3) \\
\end{align}\]
By comparing (3) and the general equation, we get
$a={{l}_{1}}{{l}_{2}};b={{m}_{1}}{{m}_{2}};c={{n}_{1}}{{n}_{2}};2h=({{l}_{1}}{{m}_{2}}+{{l}_{2}}{{m}_{1}});2f=({{m}_{1}}{{n}_{2}}+{{n}_{1}}{{m}_{2}});2g=({{l}_{1}}{{n}_{2}}+{{l}_{2}}{{n}_{1}});$
Then,
\[\begin{align}
  & (2f)(2g)(2h)=({{m}_{1}}{{n}_{2}}+{{n}_{1}}{{m}_{2}})({{l}_{1}}{{n}_{2}}+{{l}_{2}}{{n}_{1}})({{l}_{1}}{{m}_{2}}+{{l}_{2}}{{m}_{1}}) \\
 & \Rightarrow ({{m}_{1}}{{n}_{2}}+{{n}_{1}}{{m}_{2}})\left( {{l}_{1}}^{2}{{n}_{2}}{{m}_{2}}+{{l}_{1}}{{l}_{2}}{{m}_{1}}{{n}_{2}}+{{l}_{1}}{{l}_{2}}{{n}_{1}}{{m}_{2}}+{{l}_{2}}^{2}{{m}_{1}}{{n}_{1}} \right) \\
 & \Rightarrow {{l}_{1}}^{2}{{n}_{2}}^{2}{{m}_{2}}{{m}_{1}}+{{l}_{1}}{{l}_{2}}{{m}_{1}}^{2}{{n}_{2}}^{2}+{{l}_{1}}{{l}_{2}}{{m}_{1}}{{m}_{2}}{{n}_{1}}{{n}_{2}}+{{l}_{2}}^{2}{{m}_{1}}^{2}{{n}_{1}}{{n}_{2}}+{{l}_{1}}^{2}{{m}_{2}}^{2}{{n}_{1}}{{n}_{2}}+{{l}_{1}}{{l}_{2}}{{m}_{1}}{{m}_{2}}{{n}_{1}}{{n}_{2}}+{{l}_{1}}{{l}_{2}}{{n}_{1}}^{2}{{m}_{2}}^{2}+{{l}_{2}}^{2}{{n}_{1}}^{2}{{m}_{1}}{{m}_{2}} \\
 & \Rightarrow {{l}_{1}}{{l}_{2}}({{m}_{1}}^{2}{{n}_{2}}^{2}+{{n}_{1}}^{2}{{m}_{2}}^{2})+{{m}_{1}}{{m}_{2}}({{l}_{1}}^{2}{{n}_{2}}^{2}+{{l}_{2}}^{2}{{n}_{1}}^{2})+{{n}_{1}}{{n}_{2}}({{l}_{2}}^{2}{{m}_{1}}^{2}+{{l}_{1}}^{2}{{m}_{2}}^{2})+2{{l}_{1}}{{l}_{2}}{{m}_{1}}{{m}_{2}}{{n}_{1}}{{n}_{2}} \\
\end{align}\]
 On simplifying, we get
\[\begin{align}
  & 8fgh={{l}_{1}}{{l}_{2}}\left[ {{\left( {{m}_{1}}{{n}_{2}}+{{n}_{1}}{{m}_{2}} \right)}^{2}}-2{{m}_{1}}{{m}_{2}}{{n}_{1}}{{n}_{2}} \right]+{{m}_{1}}{{m}_{2}}\left[ {{\left( {{l}_{1}}{{n}_{2}}+{{l}_{2}}{{n}_{1}} \right)}^{2}}-2{{l}_{1}}{{n}_{2}}{{l}_{2}}{{n}_{1}} \right] \\
 & +{{n}_{1}}{{n}_{2}}\left[ {{\left( {{l}_{2}}{{m}_{1}}+{{l}_{1}}{{m}_{2}} \right)}^{2}}-2{{l}_{2}}{{m}_{1}}{{l}_{1}}{{m}_{2}} \right]+2{{l}_{1}}{{l}_{2}}{{m}_{1}}{{m}_{2}}{{n}_{1}}{{n}_{2}} \\
 & \Rightarrow a\left[ {{(2f)}^{2}}-2bc \right]+b\left[ {{(2g)}^{2}}-2ac \right]+c\left[ {{(2h)}^{2}}-2ab \right]+2abc \\
 & \Rightarrow 4a{{f}^{2}}-2abc+4b{{g}^{2}}-2abc+4b{{g}^{2}}-2abc+2abc \\
 & \Rightarrow -4abc+4a{{f}^{2}}+4b{{g}^{2}}+4b{{g}^{2}} \\
\end{align}\]
\[\begin{align}
  & 8fgh=-4abc+4a{{f}^{2}}+4b{{g}^{2}}+4b{{g}^{2}} \\
 & \Rightarrow 2fgh=-abc+a{{f}^{2}}+b{{g}^{2}}+b{{g}^{2}} \\
 & \Rightarrow abc+afgh-a{{f}^{2}}-b{{g}^{2}}-b{{g}^{2}}=0 \\
 & \therefore \Delta =0 \\
\end{align}\]
Thus, the given equation represents pair of straight lines.
If ${{h}^{2}}=ab$, the angle between these two lines becomes zero. I.e.,
$\cos \theta =\dfrac{a+b}{\sqrt{{{(a-b)}^{2}}+4{{h}^{2}}}}$
Substituting ${{h}^{2}}=ab$ in the above angle, we get
$\begin{align}
  & \cos \theta =\dfrac{a+b}{\sqrt{{{(a-b)}^{2}}+4(ab)}} \\
 & \text{ }=\dfrac{a+b}{a+b} \\
 & \text{ }=1 \\
 & \Rightarrow \theta ={{\cos }^{-1}}(1)=0 \\
\end{align}$
Since the angle between them is zero, these two lines are parallel.



Option ‘A’ is correct



Note: To solve such a problem, we need to assume two lines and solve them in the above process to prove them to be a pair of parallel lines.