
The equation ${{x}^{2}}+k{{y}^{2}}+4xy=0$ represents two coincident lines, if $k=$
A. $0$
B. $1$
C. $4$
D. $16$
Answer
163.2k+ views
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.
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:
$ax+hy\pm y\sqrt{{{h}^{2}}-ab}=0$
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.
Recently Updated Pages
Geometry of Complex Numbers – Topics, Reception, Audience and Related Readings

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

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

Instantaneous Velocity - Formula based Examples for JEE

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

JEE Advanced 2025 Notes

JEE Main Chemistry Question Paper with Answer Keys and Solutions

Total MBBS Seats in India 2025: Government and Private Medical Colleges

NEET Total Marks 2025
