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Question
Find the differential equation all parabolas having a length of latus rectum 4a and axis is parallel to the axis.
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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.
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