Question

Form the differential equation of all parabolas which have 4b as latus rectum and whose axis is parallel to the Y-axis.

Answer 1

Let A(h, k) be the vertex of the parabola which has 4b as a latus rectum and whose axis is parallel to Y-axis. Then the equation of the parabola is

(x - h)² = 4b(y - k)    .....(1)

where h and k are arbitrary constants.

Differentiating both sides of (1) w.r.t. x, we get

`2("x - h") * "d"/"dx" ("x - h") = "4b""d"/"dx" ("y - k")`

∴ `2("x - h") xx (1 - 0) = "4b"("dy"/"dx" - 0)`

∴ (x - h) = 2b`"dy"/"dx"`

Differentiating again w.r.t. x, we get

`1 - 0 = "2b"("d"^2"y")/"dx"^2`

∴ `"2b"("d"^2"y")/"dx"^2 - 1 = 0`

This is the required D.E.