
What is the equation of the line which cuts off the intercepts $2a\sec \theta $ and $2a\csc \theta $?
A. $x\sin \theta + y\cos \theta - 2a = 0$
B. $x\cos \theta + y\sin \theta - 2a = 0$
C. $x\sec \theta + y\csc \theta - 2a = 0$
D. $x\csc \theta + y\sec \theta - 2a = 0$
Answer
232.8k+ views
Hint: We will apply the intercepted form of the equation to find the equation of the line. Then apply trigonometry ratios to simplify the equation.
Formula Used:
The intercept form of a line is $\dfrac{x}{a} + \dfrac{y}{b} = 1$, where $a$ is the x-intercept and $b$ is the y-intercept.
Trigonometry ratio:
$\sec \theta = \dfrac{1}{{\cos \theta }}$
$\csc \theta = \dfrac{1}{{\sin \theta }}$
Complete step by step solution:
Given a line that cuts off the intercepts $2a\sec \theta $ and $2a\csc \theta $.
The x-intercept of the line is $2a\sec \theta $.
So, the value of $a$ is $2a\sec \theta $.
The y-intercept of the line is $2a\csc \theta $.
So, the value of $b$ is $2a\csc \theta $.
Now put the value of $a$ and $b$ in the intercept form of a line.
$\dfrac{x}{{2a\sec \theta }} + \dfrac{y}{{2a\csc \theta }} = 1$
Take common $\dfrac{1}{{2a}}$ from the left side of the equation
$ \Rightarrow \dfrac{1}{{2a}}\left[ {\dfrac{x}{{\sec \theta }} + \dfrac{y}{{\csc \theta }}} \right] = 1$
Multiply both sides by $2a$
$ \Rightarrow 2a \cdot \dfrac{1}{{2a}}\left[ {\dfrac{x}{{\sec \theta }} + \dfrac{y}{{\csc \theta }}} \right] = 1 \cdot 2a$
Cancel out $2a$ from the left side
$ \Rightarrow \dfrac{x}{{\sec \theta }} + \dfrac{y}{{\csc \theta }} = 2a$
Apply the formula of trigonometry ratio formula
$ \Rightarrow \dfrac{x}{{\dfrac{1}{{\cos \theta }}}} + \dfrac{y}{{\dfrac{1}{{\sin \theta }}}} = 2a$
Simplify the above equation
$ \Rightarrow x\cos \theta + y\sin \theta = 2a$
Subtract $2a$ from both sides
$ \Rightarrow x\cos \theta + y\sin \theta - 2a = 2a - 2a$
$ \Rightarrow x\cos \theta + y\sin \theta - 2a = 0$
Option ‘B’ is correct
Note: To this type, you need to know the slope intercepts form of a line that is $\dfrac{x}{a} + \dfrac{y}{b} = 1$ where a is the x-intercept and b is the y-intercept. By substituting the value of a and b, you get the required answer.
Formula Used:
The intercept form of a line is $\dfrac{x}{a} + \dfrac{y}{b} = 1$, where $a$ is the x-intercept and $b$ is the y-intercept.
Trigonometry ratio:
$\sec \theta = \dfrac{1}{{\cos \theta }}$
$\csc \theta = \dfrac{1}{{\sin \theta }}$
Complete step by step solution:
Given a line that cuts off the intercepts $2a\sec \theta $ and $2a\csc \theta $.
The x-intercept of the line is $2a\sec \theta $.
So, the value of $a$ is $2a\sec \theta $.
The y-intercept of the line is $2a\csc \theta $.
So, the value of $b$ is $2a\csc \theta $.
Now put the value of $a$ and $b$ in the intercept form of a line.
$\dfrac{x}{{2a\sec \theta }} + \dfrac{y}{{2a\csc \theta }} = 1$
Take common $\dfrac{1}{{2a}}$ from the left side of the equation
$ \Rightarrow \dfrac{1}{{2a}}\left[ {\dfrac{x}{{\sec \theta }} + \dfrac{y}{{\csc \theta }}} \right] = 1$
Multiply both sides by $2a$
$ \Rightarrow 2a \cdot \dfrac{1}{{2a}}\left[ {\dfrac{x}{{\sec \theta }} + \dfrac{y}{{\csc \theta }}} \right] = 1 \cdot 2a$
Cancel out $2a$ from the left side
$ \Rightarrow \dfrac{x}{{\sec \theta }} + \dfrac{y}{{\csc \theta }} = 2a$
Apply the formula of trigonometry ratio formula
$ \Rightarrow \dfrac{x}{{\dfrac{1}{{\cos \theta }}}} + \dfrac{y}{{\dfrac{1}{{\sin \theta }}}} = 2a$
Simplify the above equation
$ \Rightarrow x\cos \theta + y\sin \theta = 2a$
Subtract $2a$ from both sides
$ \Rightarrow x\cos \theta + y\sin \theta - 2a = 2a - 2a$
$ \Rightarrow x\cos \theta + y\sin \theta - 2a = 0$
Option ‘B’ is correct
Note: To this type, you need to know the slope intercepts form of a line that is $\dfrac{x}{a} + \dfrac{y}{b} = 1$ where a is the x-intercept and b is the y-intercept. By substituting the value of a and b, you get the required answer.
Recently Updated Pages
JEE Main 2023 April 6 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 6 Shift 2 Question Paper with Answer Key

JEE Main 2023 (January 31 Evening Shift) Question Paper with Solutions [PDF]

JEE Main 2023 January 30 Shift 2 Question Paper with Answer Key

JEE Main 2023 January 25 Shift 1 Question Paper with Answer Key

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Key

Trending doubts
JEE Main 2026: Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Understanding the Electric Field of a Uniformly Charged Ring

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 12 Limits and Derivatives (2025-26)

NCERT Solutions For Class 11 Maths Chapter 10 Conic Sections (2025-26)

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Derivation of Equation of Trajectory Explained for Students

Understanding Electromagnetic Waves and Their Importance

