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 θ + y sin θ = p to identify the angle θ.

Step 2: Key Formula or Approach:

1. Equation of perpendicular line: 12x + 5y + k = 0.
2. Distance from origin: |k| / √(a² + b²) = 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 ⇒ y = -k/5
For intercept > 0:
-k/5 > 0 ⇒ k < 0

Condition 2: Distance from origin is 2.
|k| / √(12² + 5²) = 2
|k| / 13 = 2
|k| = 26
Since k < 0:
k = -26

Equation of L:
12x + 5y - 26 = 0
12x + 5y = 26

Find Angle θ:
Convert to normal form by dividing by 13:
(12/13)x + (5/13)y = 2

Comparing with x cos θ + y sin θ = p:
cos θ = 12/13
sin θ = 5/13

Since both are positive, θ lies in Quadrant I.
tan θ = 5/12
cot θ = 12/5

Calculate Expression:
tan θ + cot θ = 5/12 + 12/5
= (25 + 144) / 60
= 169/60

Step 4: Final Answer:

The value is 169/60.