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

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प्रश्न

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

योग

उत्तर

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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Formation of Differential Equations
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 6: Differential Equations - Exercise 6.2 [पृष्ठ १९६]

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बालभारती Mathematics and Statistics 2 (Arts and Science) [English] 12 Standard HSC Maharashtra State Board
अध्याय 6 Differential Equations
Exercise 6.2 | Q 3 | पृष्ठ १९६

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