
If the ratio of gradients of the lines represented by \[a{{x}^{2}}+2hxy+b{{y}^{2}}=0\] is \[1:3\], then the value of the ratio \[{{h}^{2}}:ab\] is
A. $\dfrac{1}{3}$
B. $\dfrac{3}{4}$
C. $\dfrac{4}{3}$
D. 1
Answer
216.3k+ views
Hint: The gradient is the measure of the steepness of a line. If the two lines represented by the homogeneous equation \[a{{x}^{2}}+2hxy+b{{y}^{2}}=0\] be \[y={{m}_{1}}x\] and \[y={{m}_{2}}x\], where \[{{m}_{1}}\] and \[{{m}_{2}}\] be the slope or gradient of the line. The expression \[a{{x}^{2}}+2hxy+b{{y}^{2}}\] should be comparable with the product \[\left( y-{{m}_{1}}x \right)\left( y-{{m}_{2}}x \right)={{y}^{2}}-({{m}_{1}}+{{m}_{2}})xy+{{m}_{1}}{{m}_{2}}{{x}^{2}}\]. Therefore, we must have \[{{m}_{1}}+{{m}_{2}}=\dfrac{-2h}{b}\] and \[{{m}_{1}}\cdot {{m}_{2}}=\dfrac{a}{b}\].
Formula used:
The equation of lines \[y={{m}_{1}}x\] and \[y={{m}_{2}}x\]
Complete step-by-step solution:
Let \[y={{m}_{1}}x\] and \[y={{m}_{2}}x\] be the equation of lines
Given, \[a{{x}^{2}}+2hxy+b{{y}^{2}}=0\] be the equation of pair of lines and their ratio of gradient is \[{{m}_{1}}:{{m}_{2}}=1:3\].
\[\Rightarrow \,\,\,\dfrac{{{m}_{1}}}{{{m}_{2}}}=\dfrac{1}{3}\]
\[\Rightarrow \,\,\,{{m}_{2}}=3{{m}_{1}}\] -------- (1)
Then, Sum of two slopes i.e., \[{{m}_{1}}+{{m}_{2}}=\dfrac{-2h}{b}\]
\[\Rightarrow \,\,\,{{m}_{1}}+3{{m}_{1}}=\dfrac{-2h}{b}\] (By (1))
\[\Rightarrow \,\,\,4{{m}_{1}}=\dfrac{-2h}{b}\]
\[\Rightarrow \,\,\,{{m}_{1}}=\dfrac{-2h}{4b}\]
\[\Rightarrow \,\,\,{{m}_{1}}=\dfrac{-h}{2b}\] -------- (2)
Product of two slopes i.e., \[{{m}_{1}}\cdot {{m}_{2}}=\dfrac{a}{b}\]
\[\Rightarrow \,\,\,{{m}_{1}}\cdot 3{{m}_{1}}=\dfrac{a}{b}\] (By (1))
\[\Rightarrow \,\,\,3m_{1}^{2}=\dfrac{a}{b}\] -------- (3)
Substitute the value of (2) in (3), then
\[\Rightarrow \,\,\,3{{\left( \dfrac{-h}{2b} \right)}^{2}}=\dfrac{a}{b}\]
\[\Rightarrow \,\,\,\dfrac{3{{h}^{2}}}{4{{b}^{2}}}=\dfrac{a}{b}\]
\[\therefore \,\,\,\dfrac{{{h}^{2}}}{ab}=\dfrac{4}{3}\]
Hence, the required ratio of \[{{h}^{2}}:ab\] is \[4:3\].
So the correct answer is option(C)
Note: Gradient of the line is the ratio of the change in the y-axis to the change in the x-axis, the slope m represents the gradient of the line. If two lines are perpendicular then the product of the gradient of line is always equal to -1 i.e., \[{{m}_{1}}\cdot {{m}_{2}}=-1\] and if two lines are parallel the gradient is equal in value i.e., \[{{m}_{1}}={{m}_{2}}\].
Formula used:
The equation of lines \[y={{m}_{1}}x\] and \[y={{m}_{2}}x\]
Complete step-by-step solution:
Let \[y={{m}_{1}}x\] and \[y={{m}_{2}}x\] be the equation of lines
Given, \[a{{x}^{2}}+2hxy+b{{y}^{2}}=0\] be the equation of pair of lines and their ratio of gradient is \[{{m}_{1}}:{{m}_{2}}=1:3\].
\[\Rightarrow \,\,\,\dfrac{{{m}_{1}}}{{{m}_{2}}}=\dfrac{1}{3}\]
\[\Rightarrow \,\,\,{{m}_{2}}=3{{m}_{1}}\] -------- (1)
Then, Sum of two slopes i.e., \[{{m}_{1}}+{{m}_{2}}=\dfrac{-2h}{b}\]
\[\Rightarrow \,\,\,{{m}_{1}}+3{{m}_{1}}=\dfrac{-2h}{b}\] (By (1))
\[\Rightarrow \,\,\,4{{m}_{1}}=\dfrac{-2h}{b}\]
\[\Rightarrow \,\,\,{{m}_{1}}=\dfrac{-2h}{4b}\]
\[\Rightarrow \,\,\,{{m}_{1}}=\dfrac{-h}{2b}\] -------- (2)
Product of two slopes i.e., \[{{m}_{1}}\cdot {{m}_{2}}=\dfrac{a}{b}\]
\[\Rightarrow \,\,\,{{m}_{1}}\cdot 3{{m}_{1}}=\dfrac{a}{b}\] (By (1))
\[\Rightarrow \,\,\,3m_{1}^{2}=\dfrac{a}{b}\] -------- (3)
Substitute the value of (2) in (3), then
\[\Rightarrow \,\,\,3{{\left( \dfrac{-h}{2b} \right)}^{2}}=\dfrac{a}{b}\]
\[\Rightarrow \,\,\,\dfrac{3{{h}^{2}}}{4{{b}^{2}}}=\dfrac{a}{b}\]
\[\therefore \,\,\,\dfrac{{{h}^{2}}}{ab}=\dfrac{4}{3}\]
Hence, the required ratio of \[{{h}^{2}}:ab\] is \[4:3\].
So the correct answer is option(C)
Note: Gradient of the line is the ratio of the change in the y-axis to the change in the x-axis, the slope m represents the gradient of the line. If two lines are perpendicular then the product of the gradient of line is always equal to -1 i.e., \[{{m}_{1}}\cdot {{m}_{2}}=-1\] and if two lines are parallel the gradient is equal in value i.e., \[{{m}_{1}}={{m}_{2}}\].
Recently Updated Pages
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

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

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

Understanding Collisions: Types and Examples for Students

How to Convert a Galvanometer into an Ammeter or Voltmeter

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

NCERT Solutions for Class 11 Maths Chapter 10 Conic Sections

NCERT Solutions for Class 11 Maths Chapter 9 Straight Lines

NCERT Solutions For Class 11 Maths Chapter 8 Sequences And Series

Understanding Atomic Structure for Beginners

NCERT Solutions For Class 11 Maths Chapter 12 Limits And Derivatives

