Question

A differential equation representing the family of parabolas with axis parallel to y-axis and whose length of latus rectum is the distance of the point (2, -3) form the line 3x + 4y = 5, is given by :

• A. \(11 \frac{d^2x}{dy^2} = 10\)
• B. \(10 \frac{d^2y}{dx^2} = 11\)
• C. \(11 \frac{d^2y}{dx^2} = 10\)
• D. \(10 \frac{d^2x}{dy^2} = 11\)

Hint:

To form a differential equation from a family of curves, the number of times you need to differentiate is equal to the number of arbitrary constants (parameters) in the general equation of the family. The goal is to eliminate all the parameters.

Answer 1

The Correct Option is C

Step 1: Find the length of the latus rectum.
The length of the latus rectum (LLR) is the perpendicular distance from the point (2, -3) to the line 3x + 4y - 5 = 0. Using the distance formula \(d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2+B^2}}\): \[ LLR = \frac{|3(2) + 4(-3) - 5|}{\sqrt{3^2 + 4^2}} = \frac{|6 - 12 - 5|}{\sqrt{9+16}} = \frac{|-11|}{\sqrt{25}} = \frac{11}{5} \] Step 2: Write the general equation of the family of parabolas.
The parabolas have their axis parallel to the y-axis. The general equation for such a parabola is: \[ (x-h)^2 = 4a(y-k) \] where (h, k) is the vertex and \(4a\) is the length of the latus rectum. We found that \(LLR = 4a = 11/5\). So, the equation of the family is: \[ (x-h)^2 = \frac{11}{5}(y-k) \] Here, h and k are arbitrary constants (parameters). Step 3: Form the differential equation.
Since there are two parameters (h and k), we need to differentiate the equation twice to eliminate them. Differentiate with respect to x: \[ 2(x-h) = \frac{11}{5} \frac{dy}{dx} \] This eliminates k. Now we differentiate again to eliminate h. Differentiate with respect to x again: \[ 2(1) = \frac{11}{5} \frac{d^2y}{dx^2} \] \[ 2 = \frac{11}{5} \frac{d^2y}{dx^2} \] Rearranging the terms: \[ 10 = 11 \frac{d^2y}{dx^2} \] This is the required differential equation.