
The pair of straight lines perpendicular to the pair of lines $a{{x}^{2}}+2hxy+b{{y}^{2}}=0$ has the equation
A. $a{{x}^{2}}-2hxy+b{{y}^{2}}=0$
B. $a{{x}^{2}}+2hxy+b{{y}^{2}}n=0$
C. $a{{x}^{2}}+2hxy-b{{y}^{2}}=0$
D. $a{{y}^{2}}-2hxy+b{{x}^{2}}=0$
Answer
162.3k+ views
Hint: If two pairs of straight line is perpendicular to each other then the lines of one pair of straight line is perpendicular to the lines of another pairs of straight line. The slope of one set of lines is the inverse of the slopes of another set of line.
Formula Used: 3. $y=mx+c$
4. $a{{x}^{2}}+2hxy+b{{y}^{2}}$
Complete step by step solution: The given equation of the straight line is $a{{x}^{2}}+2hxy+b{{y}^{2}}=0$.
Let the lines of the given pair of straight lines are given as-
$ y={{m}_{1}}x$
$ y={{m}_{2}}x$
Where ${{m}_{1}},{{m}_{2}}$ are the slope of the lines.
Now we know that the sum of the two slopes which is equal to $\dfrac{-2h}{b}$ and multipication of the two slopes which is equal to $\dfrac{a}{b}$ . So, we can write ${{m}_{1}}+{{m}_{2}}=\dfrac{-2h}{b}$ and ${{m}_{1}}{{m}_{2}}=\dfrac{a}{b}$ .
Now if another pair of straight lines is perpendicular to $a{{x}^{2}}+2hxy+b{{y}^{2}}=0$ then the lines are also perpendicular to the above lines.
So, the equation of the lines perpendicular to
$ y={{m}_{1}}x$
$y={{m}_{2}}x $
is given as follows-
$ y+\dfrac{1}{{{m}_{1}}}x=0 $
$y+\dfrac{1}{{{m}_{2}}}x=0 $
Thus the equation of the pair of straight lines is given as follows-
$(y+\dfrac{1}{{{m}_{1}}}x)(y+\dfrac{1}{{{m}_{2}}}x)=0 $
${{y}^{2}}+\dfrac{1}{{{m}_{1}}}xy+\dfrac{1}{{{m}_{2}}}xy+\dfrac{1}{{{m}_{1}}}.\dfrac{1}{{{m}_{2}}}{{x}^{2}}=0 $
$ {{m}_{1}}{{m}_{2}}{{y}^{2}}+xy\left( {{m}_{1}}+{{m}_{2}} \right)+{{x}^{2}}=0$
Putting the value of ${{m}_{1}}+{{m}_{2}}=\dfrac{-2h}{b}$ and ${{m}_{1}}{{m}_{2}}=\dfrac{a}{b}$ we get-
$\dfrac{a}{b}{{y}^{2}}-\dfrac{2h}{b}x+{{x}^{2}}=0$
$a{{y}^{2}}-2hxy+b{{x}^{2}}=0$
Thus we can write that the pair of straight lines perpendicular to the pair of lines $a{{x}^{2}}+2hxy+b{{y}^{2}}=0$ has the equation$a{{y}^{2}}-2hxy+b{{x}^{2}}=0$.
Option ‘D’ is correct
Note: Two straight lines are perpendicular means the angle between the straight lines is ${{90}^{o}}$ . This angle can be determined by the formula $\tan \theta =\dfrac{2\sqrt{{{h}^{2}}-ab}}{a+b}$.
Formula Used: 3. $y=mx+c$
4. $a{{x}^{2}}+2hxy+b{{y}^{2}}$
Complete step by step solution: The given equation of the straight line is $a{{x}^{2}}+2hxy+b{{y}^{2}}=0$.
Let the lines of the given pair of straight lines are given as-
$ y={{m}_{1}}x$
$ y={{m}_{2}}x$
Where ${{m}_{1}},{{m}_{2}}$ are the slope of the lines.
Now we know that the sum of the two slopes which is equal to $\dfrac{-2h}{b}$ and multipication of the two slopes which is equal to $\dfrac{a}{b}$ . So, we can write ${{m}_{1}}+{{m}_{2}}=\dfrac{-2h}{b}$ and ${{m}_{1}}{{m}_{2}}=\dfrac{a}{b}$ .
Now if another pair of straight lines is perpendicular to $a{{x}^{2}}+2hxy+b{{y}^{2}}=0$ then the lines are also perpendicular to the above lines.
So, the equation of the lines perpendicular to
$ y={{m}_{1}}x$
$y={{m}_{2}}x $
is given as follows-
$ y+\dfrac{1}{{{m}_{1}}}x=0 $
$y+\dfrac{1}{{{m}_{2}}}x=0 $
Thus the equation of the pair of straight lines is given as follows-
$(y+\dfrac{1}{{{m}_{1}}}x)(y+\dfrac{1}{{{m}_{2}}}x)=0 $
${{y}^{2}}+\dfrac{1}{{{m}_{1}}}xy+\dfrac{1}{{{m}_{2}}}xy+\dfrac{1}{{{m}_{1}}}.\dfrac{1}{{{m}_{2}}}{{x}^{2}}=0 $
$ {{m}_{1}}{{m}_{2}}{{y}^{2}}+xy\left( {{m}_{1}}+{{m}_{2}} \right)+{{x}^{2}}=0$
Putting the value of ${{m}_{1}}+{{m}_{2}}=\dfrac{-2h}{b}$ and ${{m}_{1}}{{m}_{2}}=\dfrac{a}{b}$ we get-
$\dfrac{a}{b}{{y}^{2}}-\dfrac{2h}{b}x+{{x}^{2}}=0$
$a{{y}^{2}}-2hxy+b{{x}^{2}}=0$
Thus we can write that the pair of straight lines perpendicular to the pair of lines $a{{x}^{2}}+2hxy+b{{y}^{2}}=0$ has the equation$a{{y}^{2}}-2hxy+b{{x}^{2}}=0$.
Option ‘D’ is correct
Note: Two straight lines are perpendicular means the angle between the straight lines is ${{90}^{o}}$ . This angle can be determined by the formula $\tan \theta =\dfrac{2\sqrt{{{h}^{2}}-ab}}{a+b}$.
Recently Updated Pages
JEE Advanced 2021 Physics Question Paper 2 with Solutions

Solutions Class 12 Notes JEE Advanced Chemistry [PDF]

JEE Advanced 2022 Chemistry Question Paper 2 with Solutions

JEE Advanced 2025 Revision Notes for Chemistry Energetics - Free PDF Download

JEE Advanced Marks vs Rank 2025 - Predict IIT Rank Based on Score

JEE Advanced 2022 Maths Question Paper 2 with Solutions

Trending doubts
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

IIT CSE Cutoff: Category‐Wise Opening and Closing Ranks

Top IIT Colleges in India 2025

IIT Fees Structure 2025

IIT Roorkee Average Package 2025: Latest Placement Trends Updates

Other Pages
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

NCERT Solutions for Class 11 Maths Chapter 4 Complex Numbers and Quadratic Equations
