I am trying to solve the following question: A gun fires a shot from $O$ with initial speed $V$ at an angle $\theta$ with the horizontal.If the acceleration due to gravity is constant ($g$) prove that the shot describes a parabola of focal length: $$\frac{V^2\cos^2\theta}{2g}$$
If the initial speed $V$ is fixed but the direction of firing can be varied prove that the region of vulnerability (i.e. the set of points that can be hit) consists of points within and on the paraboloid whose equation (referred to a cartesian $x, y, z$-frame with origin at $O$ and $z$-axis vertically upwards) is:
$$x^2+y^2+\frac{2V^2}{g}z=\frac{V^4}{g^2}$$
My attempt:
Taking $\displaystyle\frac{d^2x}{dt^2}=0$ and $\displaystyle\frac{d^2y}{dt^2}=-g$, we use the initial conditions $\displaystyle t=0, x=0, \frac{dx}{dt}=V\cos{\theta}, y=0, \frac{dy}{dt}=V\sin{\theta}$ to derive:
$$x=V\cos{\theta}t, y=-\frac{gt^2}{2}+V\sin{\theta}t$$
Reducing this to $y$ as a function of $x$ gives $$y=-\frac{g\sec^2{\theta}}{2V^2}x^2+x\tan{\theta}\Longrightarrow \left(x-\frac{V^2}{g\sec^2{\theta}}\tan{\theta}\right)^2=4\frac{V^2}{2g\sec^2{\theta}}\left(y-\frac{g\sec{\theta}}{2V}\sqrt{\frac{\tan{\theta}}{g}}\right)$$
This is of the form $(x-p)^2=4a(y-q)$ with vertex $(p,q)$ and focal length $a$. $$\therefore \text{focal length}=\frac{V^2}{2g\sec^2{\theta}}=\frac{V^2\cos^2\theta}{2g}$$
Then, taking the cartesian equation I wish it to be an equation in terms if $\tan\theta$. $$\therefore y=-\frac{g\sec^2{\theta}}{2V^2}x^2+x\tan{\theta}\Longrightarrow \frac{gx^2}{2V^2}\tan^2{\theta}-x\tan\theta +\frac{gx^2}{2V^2}+y=0$$
To find the parabola within and on which the bullets hit as $\theta$ varies, I want only one solution to this equation since every point within this parabola will have two solutions and every point outside will have none i.e I want $\Delta=b^2-4ac=0$. $$\therefore x^2=4\frac{gx^2}{2V^2}\left(\frac{gx^2}{2V^2}+y\right)\Longrightarrow x^2+\frac{2V^2}{g}y=\frac{V^4}{g^2}.$$
This looks similar to the required curve, however I don't know how I can transform it into the required three dimensional curve. Could someone please help me with this?
I think that you cannot transform it into the required three dimensional curve.
Instead, we can start with the followings : $$\dfrac{d^2x}{dt^2}=0,\quad \dfrac{d^2y}{dt^2}=0,\quad \dfrac{d^2z}{dt^2}=-g$$ with the initial conditions $x=y=z=0$ and $$\frac{dx}{dt}=V\sin\theta\cos\varphi,\quad \frac{dy}{dt}=V\sin\theta\sin\varphi,\quad \frac{dz}{dt}=V\cos\theta$$ where $\theta$ is the polar angle, and $\varphi$ is the azimuthal angle (see here).
From these, we get $$x=Vt\sin\theta\cos\varphi,\quad y=Vt\sin\theta\sin\varphi,\quad z=Vt\cos\theta-\frac 12gt^2$$
from which we have $$x^2+y^2=V^2t^2\sin^2\theta$$ and $$x^2+y^2+\bigg(z+\frac 12gt^2\bigg)^2=V^2t^2$$
Seeing this as a quadratic equation of $t^2$, we get $$\frac{g^2}{4}(t^2)^2+(gz-V^2)t^2+x^2+y^2+z^2=0$$
Finally, $\Delta=0$ gives $$(gz-V^2)^2-4\cdot\frac{g^2}{4}(x^2+y^2+z^2)=0$$ i.e. $$x^2+y^2+\frac{2V^2}{g}z=\frac{V^4}{g^2}$$