Question

The differential equation of all lines where the length of the normal from the origin is p and the inclination of the normal is \(α\) is... (where p and \(α\) are arbitrary constants)

• A. \(\frac{d^2y}{dx^2} = 0\)
• B. \(\frac{dy}{dx} = 0\)
• C. \(\frac{dy}{dx} = -cotα\)
• D. \(\frac{d^2y}{dx^2} = csc^2α\)

Hint:

The family has two arbitrary constants, so the equation is of second order.

Answer 1

The Correct Option is A

Step 1: Understand the concept
A line whose normal from the origin has length \(p\) and inclination \(\alpha\) has the equation \(x\cos\alpha + y\sin\alpha = p\). Here both \(p\) and \(\alpha\) are arbitrary constants, so the family is all straight lines (two parameters).

Step 2: Solve for y
\[ y = -x\cot\alpha + p\csc\alpha \]
This is of the form \(y = mx + c\) with two arbitrary constants \(m = -\cot\alpha\) and \(c = p\csc\alpha\).

Step 3: Eliminate the constants
Differentiate once: \(\frac{dy}{dx} = -\cot\alpha\). Differentiate again: \(\frac{d^2y}{dx^2} = 0\). Since two constants must be eliminated, we need the second order equation.

Step 4: Check the options
Option (B) \(\frac{dy}{dx} = 0\) is only the horizontal lines, and option (C) fixes the slope, which is not arbitrary. Option (D) is not zero for a straight line. So the answer is \(\frac{d^2y}{dx^2} = 0\), option (A).

Final Answer:
The differential equation is y'' = 0. This is option (A). \[ \boxed{\text{(A) }\frac{d^2y}{dx^2}=0} \]