For what values of $p$, ${{y}^{2}}+xy+p{{x}^{2}}-x-2y=0$ represents two straight lines
A. $2$
B. $\dfrac{1}{3}$
C. $\dfrac{1}{4}$
D. $\dfrac{1}{2}$
Answer
270.9k+ views
Hint: In this question, we need to find the value of $p$in the equation that represents two straight lines. So, we can apply the formula $\Delta =0$ to find the required value.
Formula UsedThe 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 ${{h}^{2}}=ab$, 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
${{y}^{2}}+xy+p{{x}^{2}}-x-2y=0\text{ }...(1)$
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
$a=p;h=\dfrac{1}{2};b=1;g=\dfrac{-1}{2};f=-1;c=0$
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
$\begin{align}
& abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}=0 \\
& \Rightarrow (p)(1)(0)+2(-1)(\dfrac{-1}{2})(\dfrac{1}{2})-p{{\left( -1 \right)}^{2}}-(1){{\left( \dfrac{-1}{2} \right)}^{2}}-(0){{\left( \dfrac{1}{2} \right)}^{2}}=0 \\
& \Rightarrow 0+\dfrac{1}{2}-p-\dfrac{1}{4}-0=0 \\
& \therefore p=\dfrac{1}{4} \\
\end{align}$
Thus, the value is $p=\dfrac{1}{4}$.
Option ‘C’ 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 UsedThe 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 ${{h}^{2}}=ab$, 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
${{y}^{2}}+xy+p{{x}^{2}}-x-2y=0\text{ }...(1)$
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
$a=p;h=\dfrac{1}{2};b=1;g=\dfrac{-1}{2};f=-1;c=0$
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
$\begin{align}
& abc+2fgh-a{{f}^{2}}-b{{g}^{2}}-c{{h}^{2}}=0 \\
& \Rightarrow (p)(1)(0)+2(-1)(\dfrac{-1}{2})(\dfrac{1}{2})-p{{\left( -1 \right)}^{2}}-(1){{\left( \dfrac{-1}{2} \right)}^{2}}-(0){{\left( \dfrac{1}{2} \right)}^{2}}=0 \\
& \Rightarrow 0+\dfrac{1}{2}-p-\dfrac{1}{4}-0=0 \\
& \therefore p=\dfrac{1}{4} \\
\end{align}$
Thus, the value is $p=\dfrac{1}{4}$.
Option ‘C’ 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
Geometry of Complex Numbers Explained

JEE General Topics in Chemistry Important Concepts and Tips

JEE Extractive Metallurgy Important Concepts and Tips for Exam Preparation

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

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

Electricity and Magnetism Explained: Key Concepts & Applications

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

JEE Main Participating Colleges 2026 - A Complete List of Top Colleges

Understanding the Electric Field of a Uniformly Charged Ring

Understanding Atomic Structure for Beginners

Derivation of Equation of Trajectory Explained for Students

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

Other Pages
JEE Advanced 2026 Notification Out with Exam Date, Registration (Extended), Syllabus and More

JEE Advanced Percentile vs Marks 2026: JEE Main Cutoff, AIR & IIT Admission Guide

JEE Advanced Weightage Chapter Wise 2026 for Physics, Chemistry, and Mathematics

How to Convert a Galvanometer into an Ammeter or Voltmeter

Electron Gain Enthalpy and Electron Affinity Explained

Understanding Electromagnetic Waves and Their Importance

