The differential equation of the family of pa | vedclass

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

Vedclass · Accepted Answer

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

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(C) The equation of the family of parabolas with vertex at $(0,-1)$ and axis along the $Y$-axis is given by $x^2 = 4a(y+1)$ $(i)$. On differentiating both sides with respect to $x$, we get $2x = 4a y^{\prime}$, which implies $a = \frac{x}{2y^{\prime}}$. Substituting the value of $a$ into equation $(i)$, we get $x^2 = 4 \left( \frac{x}{2y^{\prime}} \right) (y+1)$. Simplifying this, we get $x = \frac{2(y+1)}{y^{\prime}}$. Thus, $x y^{\prime} = 2y + 2$, or $x y^{\prime} - 2y - 2 = 0$.

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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