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