Question

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

• A. $y{y}^{\prime }+2xy+1=0$
• B. $x{y}^{\prime }+y+1=0$
• C. $x{y}^{\prime }+2y+2=0$
• D. $x{y}^{\prime }-y-1=0$

Answer 1

The correct option is C $x{y}^{\prime }+2y+2=0$

In the standard equation of parabola $y^2=4ax,$ the vertex is $(0,0)$ and the axis is $y=0$

Similarly, the axis here is $x=0$ and the vertex is $(0,-1).$

The equation thus becomes $x^2=\pm 4a(y+1)$

i.e. $x^2\pm (4ay+4a)=0$

Differentiating w.r.t $x$, we get $2x\pm 4a{y}^{\prime }=0$

$\Rightarrow a=\pm \frac{2x}{4y^{\prime }}=\pm \frac{x}{2y^{\prime }}$

The equation thus becomes $x^2\pm (\frac{2yx}{y^{\prime }}+\frac{2x}{y^{\prime }})=0$

i.e. $x{y}^{\prime }\pm (2y+2)=0$