
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
162.6k+ views
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.
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 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.
Recently Updated Pages
Fluid Pressure - Important Concepts and Tips for JEE

JEE Main 2023 (February 1st Shift 2) Physics Question Paper with Answer Key

Impulse Momentum Theorem Important Concepts and Tips for JEE

Graphical Methods of Vector Addition - Important Concepts for JEE

JEE Main 2022 (July 29th Shift 1) Chemistry Question Paper with Answer Key

JEE Main 2023 (February 1st Shift 1) Physics Question Paper with Answer Key

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

JoSAA JEE Main & Advanced 2025 Counselling: Registration Dates, Documents, Fees, Seat Allotment & Cut‑offs

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

NEET 2025 – Every New Update You Need to Know

Verb Forms Guide: V1, V2, V3, V4, V5 Explained

NEET Total Marks 2025

1 Billion in Rupees
