Envelope of familiy of parabolas

Question

In the $XY-$ plane consider the family $\mathcal{P}$ of inverted parabolas that pass through the origin such that the tangent vector at the origin has equal norm (say $\lambda>0$) for all of them. One of such representation can be $$\mathcal{P}=\{p(t)=(x(t),y(t)):x(t)=\lambda\cos(\alpha)t,y(t)=\lambda\sin(\alpha)t-at^2,\alpha\in[0,\pi],a>0\}$$ Find the curve that is tangent to all of the parabolas in $\mathcal{P}$.

Answer 1

For a proper fixed value of $x$, if we vary the value of the angle, we obtain different values of $y$, i.e., if $x=\lambda\cos(\alpha)t=c_0$, then $t=\dfrac{c_0}{\lambda\cos(\alpha)}$, so considering $y$ in terms of $\alpha$ $$y(\alpha)=\lambda\sin(\alpha)\dfrac{c_0}{\lambda\cos(\alpha)}-a\left(\dfrac{c_0}{\lambda\cos(\alpha)}\right)^2=\tan(\alpha)c_0-\dfrac{ac_0^2}{\lambda^2\cos^2(\alpha)}$$

Now we look for a value of $\alpha$ such that $y(\alpha)$ is maximum. \begin{eqnarray*} y'(\alpha)&=&\sec^2(\alpha)c_0-\dfrac{2ac_0^2\sec^2(\alpha)\tan(\alpha)}{\lambda^2}\\ y'(\alpha)=0&\Longrightarrow&\sec^2(\alpha)c_0\left(\lambda^2-2ac_0\tan(\alpha)\right)=0\\ \Longrightarrow\quad\alpha_0&=&\arctan\left(\dfrac{\lambda^2}{2ac_0}\right)\\ \Longrightarrow\quad y(\alpha_0)&=&c_0\left(\dfrac{\lambda^2}{2ac_0}\right)-\dfrac{ac_0^2}{\lambda^2}\left(1+\dfrac{\lambda^4}{4a^2c_0^2}\right)\\ &=&\dfrac{\lambda^2}{2a}-\dfrac{ac_0^2}{\lambda^2}-\dfrac{\lambda^2}{4a}\\ &=&\dfrac{\lambda^2}{4a}-\dfrac{ac_0^2}{\lambda^2} \end{eqnarray*} which is the value of the curve we're looking for at any given value $c_0$ of $x$, hence the function we want is $$\varphi(x)=\dfrac{\lambda^2}{4a}-\dfrac{ax^2}{\lambda^2}$$