Find the differential equation all parabolas having a length of latus rectum 4a and axis is parallel to the axis.
Solution
Let A(h, k) be the vertex of the parabola whose length of the latus rectum is 4a.
Then the equation of the parabola is
(y - k)2 = 4a(x - h), where h and k are arbitrary constants. Differentiating w.r.t. x, we get
`2("y - k") * "d"/"dx" ("y - k") = 4"a" "d"/"dx" ("x - h")`
∴ `2("y - k")("dy"/"dx" - 0) = "4a"(1 - 0)`
∴ `2("y - k")"dy"/"dx" = "4a"`
∴ `("y - k")"dy"/"dx" = "2a"` ...(1)
Differentiating again w.r.t. x, we get
`("y - k") * "d"/"dx" ("dy"/"dx") + "dy"/"dx" * "d"/"dx" ("y - k") = 0`
∴ `("y - k")("d"^2"y")/"dx"^2 + "dy"/"dx" * ("dy"/"dx" - 0) = 0`
∴ `("y - k")("d"^2"y")/"dx"^2 + ("dy"/"dx")^2 = 0`
∴ `"2a"/(("dy"/"dx")) * ("d"^2"y")/"dx"^2 + ("dy"/"dx")^2 = 0` ....[By (1)]
∴ `"2a" ("d"^2"y")/"dx"^2 + ("dy"/"dx")^3 = 0`
This is the required D.E.
