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

Answer 1

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.