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) from the line 3x + 4y = 5, is given by :

Solution

Maths sir · 4 steps

Find the latus rectum as a number first, then differentiate twice and both h and k disappear.

1. Latus rectum length. The distance of (2, -3) from 3x + 4y - 5 = 0 is (|3(2) + 4(-3) - 5|)/√(3² + 4²) = (|-11|)/5 = 11/5. So the latus rectum is 4a = 11/5.
2. Write the family. A parabola with axis parallel to the y-axis has the form (x - h)² = 4a(y - k). Here 4a is fixed at 11/5, and h and k are the two arbitrary constants.

   (x - h)² = 11/5(y - k)

3. Differentiate twice. Differentiating with respect to x gives 2(x - h) = 11/5dy/dx, which removes k. Differentiating again gives 2 = 11/5d²y/dx², which removes h.
4. So. Multiplying through by 5 gives 11d²y/dx² = 10, which is option (d).

The idea to keep. Two arbitrary constants mean you differentiate twice. If the axis is parallel to the y-axis, x is squared, so the equation uses d²y/dx².

Answer from the NTA final answer key. Solution by PrepWiser.