The differential equation of the family of pa | vedclass

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Question

(C) The equation of the family of parabolas with vertex at $(0, -1)$ and axis along the $Y$-axis is given by:$x^2 = 4a(y + 1)$  --- $(i)$Differentiati

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$x y^{\prime}-2 y-2=0$

Vedclass · Answer

(C) The equation of the family of parabolas with vertex at $(0, -1)$ and axis along the $Y$-axis is given by:$x^2 = 4a(y + 1)$  --- $(i)$Differentiating both sides with respect to $x$,we get:$2x = 4a y^{\prime}$$a = \frac{2x}{4y^{\prime}} = \frac{x}{2y^{\prime}}$Substituting the value of $a$ into equation $(i)$:$x^2 = 4 \left( \frac{x}{2y^{\prime}} ight) (y + 1)$$x^2 = \frac{2x}{y^{\prime}} (y + 1)$Assuming $x 
eq 0$,we divide by $x$:$x = \frac{2(y + 1)}{y^{\prime}}$$x y^{\prime} = 2y + 2$$x y^{\prime} - 2y - 2 = 0$

The differential equation of the family of parabolas with vertex at $(0, -1)$ and having axis along the $Y$-axis is

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