Question

Find the differential equation of family of parabolas with vertex at $(h,0)$ and the principal axis along the $x$-axis.

Answer 1

$PS=\sqrt{{\{\alpha -(a+h\left)\right\}}^2+{\beta }^2}$
$PM=\left|\frac{\alpha +a-h}{\sqrt{1^2+0^2}}\right|=|\alpha +a-h|$
For parabola,
$PS=PM$
$P{S}^2=P{M}^2$
${\{\alpha -(a+h{)}^2\}}^2+{\beta }^2={\{\alpha +(a-h\left)\right\}}^2$
${\alpha }^2+(a+h{)}^2-2\alpha (a+h)+{\beta }^2={\alpha }^2+(a-h{)}^2+2\alpha (a-h)$
${\alpha }^2+a^2+h^2+2ah-2\alpha a-2\alpha h+{\beta }^2={\alpha }^2+a^2+h^2-2ah+2\alpha a-2\alpha h$
$4ah-4a\alpha +{\beta }^2=0$
On generalisation,
$4ah-4ax+y^2=0$
$y^2=4a(x-h)$ ... (1)
On differentiating,
$2y\frac{dy}{dx}=4a(1-0)$
$2y\frac{dy}{dx}=4a....\left(2\right)$
Putting the value of $4a$ in equal (1),
$y^2=2y\frac{dy}{dx}(x-h)$
$\frac{dy}{dx}=\frac{y}{2(x-h)}$.