
The equation $2{{x}^{2}}+4xy-k{{y}^{2}}+4x+2y-1=0$ represents a pair of lines. The value of $k$ is equal to
A. $\dfrac{-5}{3}$
B. $\dfrac{5}{3}$
C. $\dfrac{1}{3}$
D. $\dfrac{-1}{3}$
Answer
221.7k+ views
Hint: In this question, we need to find the value of $k$ in the equation that represents two straight lines. So, we can apply the formula $\Delta =0$ to find the required value.
Formula Used:The equation of the 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 equation is
$2{{x}^{2}}+4xy-k{{y}^{2}}+4x+2y-1=0\text{ }...(1)$
But we have the general equation of pair lines as
$S\equiv a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0\text{ }...(2)$
Comparing (1) and (2), we get
$a=2;h=2;b=-k;g=2;f=1;c=-1$
If the given equation (1) represents two pairs of lines, then
$\Delta =abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}=0\text{ }...(3)$
On substituting the above values in (3), we get
$\begin{align}
& abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}=0 \\
& \Rightarrow (2)(-k)(-1)+2(1)(2)(2)-2{{\left( 1 \right)}^{2}}-(-k){{\left( 2 \right)}^{2}}-(-1){{\left( 2 \right)}^{2}}=0 \\
& \Rightarrow 2k+8-2+4k+4=0 \\
& \Rightarrow 6k=-10 \\
& \Rightarrow k=\dfrac{-10}{6} \\
& \therefore k=\dfrac{-5}{3} \\
\end{align}$
Thus, the value is $k=\dfrac{-5}{3}$.
Option ‘A’ is correct
Note: Here, the given equation represents pair of lines. So, the given equation should satisfy the condition we have $abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}=0$. Then, by substituting the values into this condition, we get the required values. In this problem, we need to find the coefficient of ${{x}^{2}}$ in the given equation. So, the we applied above formula. On simplifying, we get the required value.
Formula Used:The equation of the 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 equation is
$2{{x}^{2}}+4xy-k{{y}^{2}}+4x+2y-1=0\text{ }...(1)$
But we have the general equation of pair lines as
$S\equiv a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0\text{ }...(2)$
Comparing (1) and (2), we get
$a=2;h=2;b=-k;g=2;f=1;c=-1$
If the given equation (1) represents two pairs of lines, then
$\Delta =abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}=0\text{ }...(3)$
On substituting the above values in (3), we get
$\begin{align}
& abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}=0 \\
& \Rightarrow (2)(-k)(-1)+2(1)(2)(2)-2{{\left( 1 \right)}^{2}}-(-k){{\left( 2 \right)}^{2}}-(-1){{\left( 2 \right)}^{2}}=0 \\
& \Rightarrow 2k+8-2+4k+4=0 \\
& \Rightarrow 6k=-10 \\
& \Rightarrow k=\dfrac{-10}{6} \\
& \therefore k=\dfrac{-5}{3} \\
\end{align}$
Thus, the value is $k=\dfrac{-5}{3}$.
Option ‘A’ is correct
Note: Here, the given equation represents pair of lines. So, the given equation should satisfy the condition we have $abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}=0$. Then, by substituting the values into this condition, we get the required values. In this problem, we need to find the coefficient of ${{x}^{2}}$ in the given equation. So, the we applied above formula. On simplifying, we get the required value.
Recently Updated Pages
States of Matter Chapter For JEE Main Chemistry

Mutually Exclusive vs Independent Events: Key Differences Explained

Area vs Volume: Key Differences Explained for Students

Conduction Explained: Definition, Examples & Science for Students

Balancing of Redox Reactions - Important Concepts and Tips for JEE

Atomic Size - Important Concepts and Tips for JEE

Trending doubts
JEE Main 2026: Application Form Open, Exam Dates, Syllabus, Eligibility & Question Papers

Derivation of Equation of Trajectory Explained for Students

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Angle of Deviation in a Prism

How to Convert a Galvanometer into an Ammeter or Voltmeter

Degree of Dissociation: Meaning, Formula, Calculation & Uses

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

Ideal and Non-Ideal Solutions Explained for Class 12 Chemistry

Understanding the Electric Field of a Uniformly Charged Ring

Understanding Atomic Structure for Beginners

Understanding Electromagnetic Waves and Their Importance

Understanding Collisions: Types and Examples for Students

