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Question
Find the differential equation of the family of all the parabolas with latus rectum 4a and whose axes are parallel to the x-axis
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Solution
Given the equation of family of parabolas with latus rectum 4a and axes are parallel to x-axis then
(y – b)2 = 4a(x – a), where (a, b) is the vertex of parabola.
y2 – 2yb + b2 = 4ax – 4a2 ........(1)
Differentiating equation (1) with respect to x, we get
`2y ("d"y)/("d"x) - 2"b" ("d"y)/("d"x) + 0 = 4"a" - 0`
`2(y ("d"y)/("d"x) - "b" ("d"y)/("d"x))` = 4a
`("d"y)/("d"x) (y - "b") = (4"a")/2`
`("d"y)/("d"x) (y - "b")` = 2a
`y ("d"y)/("d"x) - 2"a" = "b" ("d"y)/("d"x)`
∵ `("d"y)/("d"x)` = y'
∴ yy' – 2a = by' .......(2)
Differentiating equation (2) with respect to ‘x’, we get
yy”+ y’y’ = by”
yy” + y’2 = by” ……. (3)
Substituting the b value in (3), we get
yy'' + (y')2 = `((yy"'" - 2"a")/(y"'"))y"''"`
`yy"''" + (y"'")^2 - y"''" ((yy"''" - 2"a")/(y"'"))` = 0
`yy"''" + (y"'")^2 - (yy"''"y"'")/y + (2"a"y"''")/(y"'")` = 0
`yy"''" + (y"'")^2 - yy"''" + (2"a"y"''")/(y"'")` = 0
`(y"'")^2 + (2"a"y"''")/(y"'")` = 0
Multiply by y', we get
`(y"'")^3 + (2"a"y"''" xx y"'")/(y"'")` = 0
(y')3 + 2ay'' = 0
Which is a required differential equation.
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