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Question
Find the differential equation of the family of parabolas with vertex at (0, –1) and having axis along the y-axis
Solution
Equation of the family of parabolas with vertex at (0, –1) and having axis along the y-axis is
(x – 0)2 = 4a(y + 1)
x2 = 4a(y + 1) .......(1)
x2 = 4 ay + 4a
Differentiating equation (1) with respect to ‘x’, we get
2x = 4a y’
`(2x)/(y"'")` = 4a
Substituting 4a value in equation (1), we get
x² = `(2x)/(y"'") (y + 1)`
`x^2/x = 2/(y"'") (y + 1)`
x = `2/(y"'") (y + 1)`
xy’ = 2(y + 1)
xy’ = 2y + 2
xy’ – 2y – 2 = 0 is a required differential equation.
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