Question

A line L perpendicular to the line \( 5x-12y+6=0 \) makes positive intercept on the Y-axis. If the distance from the origin to the line L is 2 units and the angle made by the perpendicular drawn from the origin to the line L with positive X-axis is \( \theta \), then \( \tan \theta + \cot \theta = \)

• A. \( \frac{25}{12} \)
• B. \( \frac{625}{168} \)
• C. \( \frac{169}{60} \)
• D. \( \frac{1681}{360} \)

Hint:

When converting to normal form, always ensure the constant \( p \) on the RHS is positive. The signs of the coefficients of \( x \) and \( y \) then correctly determine the quadrant of the normal angle.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:

We first determine the equation of line L using the perpendicularity condition and the distance from the origin. Then, we convert the equation into the normal form \( x \cos \alpha + y \sin \alpha = p \) to identify the angle \( \alpha \) (which corresponds to \( \theta \) in the question).
Step 2: Key Formula or Approach:

1. Equation of perpendicular line: \( 12x + 5y + k = 0 \). 2. Distance from origin: \( \frac{|k|}{\sqrt{a^2+b^2}} = p \). 3. Normal form requires the constant term to be positive on the RHS.
Step 3: Detailed Explanation:

Given line: \( 5x - 12y + 6 = 0 \). Line L is perpendicular, so its form is \( 12x + 5y + k = 0 \). Condition 1: Positive Y-intercept. Put \( x=0 \): \( 5y = -k \implies y = -k/5 \). For intercept \( \textgreater 0 \), \( -k/5 \textgreater 0 \implies k \textless 0 \). Condition 2: Distance from origin is 2. \[ \frac{|k|}{\sqrt{12^2 + 5^2}} = 2 \implies \frac{|k|}{13} = 2 \implies |k| = 26 \] Since \( k \textless 0 \), \( k = -26 \). Equation of L: \( 12x + 5y - 26 = 0 \implies 12x + 5y = 26 \). Find Angle \( \theta \):
Convert to normal form by dividing by \( \sqrt{12^2+5^2} = 13 \): \[ \frac{12}{13}x + \frac{5}{13}y = 2 \] Comparing with \( x \cos \theta + y \sin \theta = p \): \( \cos \theta = \frac{12}{13} \) and \( \sin \theta = \frac{5}{13} \). Since both are positive, \( \theta \) is in Quadrant I. \( \tan \theta = \frac{5}{12} \), and \( \cot \theta = \frac{12}{5} \). Calculate Expression:
\[ \tan \theta + \cot \theta = \frac{5}{12} + \frac{12}{5} = \frac{25 + 144}{60} = \frac{169}{60} \]
Step 4: Final Answer:

The value is \( \frac{169}{60} \).