Consider the space of non zero real homogeneous degree $2$ polynomials in three variables upto scaling. This space is $\mathbb{R} \mathbb{P}^5$. The zero set 
of such a polynomial gives a real curve in $\mathbb{R} \mathbb{P}^2$. Let me 
define the space $X$ to be the space of real curves $[f]$ and a marked point 
$p$, such that the curve has a singularity at $p$, i.e. 
$$ X := \{ ([f], p) \in \mathbb{R} \mathbb{P}^5 \times \mathbb{R} \mathbb{P}^2: 
f(p) =0, ~~\nabla f|\_p =0 \}. $$ 


$\textbf{Question $1$:}$ Is $X$ a smooth manifold of the expected dimension (four)?


$\textbf{Question $2$:}$ If yes, is $X$ orientable or non orientable? 


$\textbf{Question $3$:} $ Is it obvious that $X \rightarrow \mathbb{RP}^2$ 
is a fiber bundle (i.e. locally trivial)? 


$\textbf{Comments on Question $3$:}$ Suppose we write a homogeneous polynomial 
explicitly as 
$$ f(X,Y,Z) := A\_1 X^2 + A\_2 Y^2 + A\_3 Z^3 + A\_4 X Y + A\_5 X Z + A\_6 YZ.$$ 
Suppose we want to trivialize the bundle near the point $[0,0,1]$. 
It is easy to see that if we ask for $f$ to have a singularity at $[0,0,1]$ 
then $A\_3, ~A\_5, ~A\_6 =0$. Let $\mathcal{U}\_{\epsilon\_1, \epsilon\_2}$ be 
an open set consisting of points of the form $[\epsilon\_1, \epsilon\_2,1]$. 
It is easy to see that the map 
$$ h: \pi^{-1} (\mathcal{U}\_{\epsilon\_1, \epsilon\_2}) 
\rightarrow \mathcal{U}\_{\epsilon\_1, \epsilon\_2} \times \mathbb{RP}^2 $$
given by 
$$ h([A\_1, \ldots, A\_6]; p):= (p, [A\_1, A\_2, A\_4]$$ 
is not a trivialization. 


Of course, that doesn't prove anything; but I am just wondering if $\pi:X \rightarrow \mathbb{RP}^2$ is a fiber bundle?


$\textbf{Comments:}$ I have two arguments giving me contradictory answers. 
First of all, note that (assuming $X$ is a manifold), 
the normal bundle to $X$ in 
$\mathbb{R} \mathbb{P}^5 \times \mathbb{R} \mathbb{P}^2$ is given by 
$$ N\_X:= \gamma\_{5}^\* \otimes \gamma\_2^{\* 2} \oplus \gamma\_5^\* \otimes T^\*\mathbb{R P}^2 \otimes \gamma\_2^{\*2} \big{|}\_X, $$ 
where $\gamma\_n$ is the tautological line bundle over $\mathbb{RP}^n$. 
(I can justify this if someone is not convinced). 


Note that 
$$ TX \approx 
\frac{T (\mathbb{R} \mathbb{P}^5 \times \mathbb{R} \mathbb{P}^2)}{N\_X}. $$
It is now easy to see that the first stiefel whitney class of the 
tangent bundle of $X$ is zero, i.e. $w\_1(TX) =0$. Hence $X$ is orientable. 


$\textbf{The above statement is incorrect:}$ One can check that 
$w\_1(TX) \neq 0$, which is consistent with the fact that $X$ is non-orientable.


On the other hand it seems to me that $~\pi: X \rightarrow \mathbb{RP}^2$ 
is a fibre bundle, with fibres $\mathbb{RP}^2$. One can show (using spectral sequences) that any $\mathbb{RP}^2$ bundle over $\mathbb{RP}^2$ is non orientable. 


$\textbf{Proof of why $X$ is a manifold:} $ Let us take $\mathbb{R}^6$ 
to be the space of real polynomials in two variables of degree at most $2$. Hence we can write such an element $\rho \in \mathbb{R}^6$ as 
$$ \rho(x,y) = \rho\_{00} + \rho\_{10} x + \rho\_{01} y + \frac{\rho\_{20}}{2} x^2 + \rho\_{11} x y + \frac{\rho\_{02}}{2} y^2. $$


Now consider the map 
$$ \psi : \mathbb{R}^6 \times \mathbb{R}^2 \rightarrow \mathbb{R}^3 $$ 
given by 
$$ \psi(\rho; x,y) := \Big( \rho\_{00} + \rho\_{10} x + \rho\_{01} y + \frac{\rho\_{20}}{2} x^2 + \rho\_{11} x y + \frac{\rho\_{02}}{2} y^2, \\
~\rho\_{10} + \rho\_{20}x + \rho\_{11} y, \\ 
~\rho\_{01} + \rho\_{11}x + \rho\_{02}y \Big). $$ 
It is easy to see that if $\psi(\rho,0,0) =0$, then the Jacobian matrix 
of $\psi(\rho,x,y)$ at $(x,y)=(0,0)$ has full rank. To see why, take the 
partial derivative of $\psi$ with respect to $\rho\_{00}$, $\rho\_{10}$ 
and $\rho\_{01}$ and plug in $(x,y) = (0,0)$. That gives us a $3\times 3$ 
identity matrix. Hence $\psi$ is transverse to the zero 
set (it is easy to see taking $(x,y) = (0,0)$ was without any loss of generality).
Hence, by Implicit Function Theorem, $\psi^{-1}(0,0,0)$ 
is a smooth manifold (even around double lines). 


Its now easy to see that the space $X$ defined will also be a manifold; 
by writing the evaluation map and the vertical derivative 
in a coordinate chart and trivialization, it reduces to the above calculation. 



Is the space of real conics with a singular point an orientable manifold?
Yes, it is a smooth manifold. No, it is not orientable.


For the first, just think geometrically, i.e., without bases: Fix a $3$-dimensional vector space $V$ and consider the homogeneous quadratic polynomials on it, which is a $6$-dimensional vector space isomorphic to $\mathsf{S}^2(V^\ast)$. Let $\mathbb{P}V$ be the projectivization of $V$, i.e., the space of $1$-dimensional subspaces of $V$. Given $L\in\mathbb{P}V$, the set of elements of $\mathsf{S}^2(V^\ast)$ that represent conics singular at $L$ is simply the $3$-dimensional space $\mathsf{S}^2(L^\perp)\subset \mathsf{S}^2(V^\ast)$, where $L^\perp\subset V^\ast$ is the space of linear functions on $V$ that vanish on $L$. In particular, the set $E\subset \mathbb{P}V\times \mathsf{S}^2(V^\ast)$ that consists of the pairs $(L,f)$ such that $f\in \mathsf{S}^2(L^\perp)$ is a smooth subbundle of rank $3$ over $\mathbb{P}V$ of the trivial bundle $\mathbb{P}V\times \mathsf{S}^2(V^\ast)$. Your space $X$ is simply $\mathbb{P}E$, the projectivization of this smooth, rank $3$ bundle. In particular, it is a smooth submanifold of $\mathbb{P}V\times \mathbb{P}\bigl(\mathsf{S}^2(V^\ast)\bigr)$ of dimension $4 = 2 + (3{-}1)$, and the projection $\mathbb{P}E\to\mathbb{P}V$ is a submersion that is a locally trivial fiber bundle.


Second, $X=\mathbb{P}E$ cannot be orientable because no projectivization of a locally trivial, rank $3$ bundle $E$ over a smooth manifold $M$ is orientable. You don't need spectral sequences to see this, you just need to produce one closed loop around which the orientation bundle is nontrivial, but these are easy to find: Just take a loop in a single fiber $\mathbb{P}E\_x$ that generates that fiber's fundamental group (which is isomorphic to $\mathbb{Z}\_2$). This is clearly an orientation-reversing loop in $\mathbb{P}E$.


**Response to Question 3:** The OP wondered why the bundle $\pi:E\to \mathbb{P}V$ is locally trivial. Here is one way to see this: Let $L\_0\in \mathbb{P}V$ be a $1$-dimensional subspace and let $X,Y,Z\in V^\ast$ be a basis of the linear functions on $V$ such that $X$ and $Y$ vanish identically on $L\_0$ (and hence are a basis of $(L\_0)^\perp$) while $z$ does not identically vanish on $L\_0$ (and hence only vanishes on $L\_0$ at $0\in L\_0$). Let $U\subset\mathbb{P}V$ be the open set consisting of those $L\in \mathbb{P}V$ such that $Z$ is not the zero linear functional when restricted to $L$. On $U$, there are two well-defined, smooth functions $x,y:U\to\mathbb{R}$ such that $x(L) = X(v)/Z(v)$ and $y(L) = Y(v)/Z(v)$ for some (and, hence, any) nonzero $v\in L$. Then, for each $L\in U$, the linear functions $X-x(L)\,Z$ and $Y - y(L)\,Z$ are a basis of $(L)^\perp\subset V^\ast$, and hence the quadratic forms 
$$
A(L) = \bigl(X-x(L)\, Z\bigr)^2,\quad 
B(L) = \bigl(X-x(L)\, Z\bigr)\bigl(Y-y(L)\, Z\bigr),\quad 
C(L) = \bigl(Y-y(L)\, Z\bigr)^2
$$
are a basis of $E\_L = \mathsf{S}^2(L^\perp)$ for each $L\in U$. Thus, they define a smooth trivialization of $\pi:E\to\mathbb{P}V$ over $U$. Obviously, $\mathbb{P}V$ can be covered by such open sets $U$, and it is easy to see that the transitions on overlaps are smooth. Thus $E$ is a smooth bundle with local smooth trivializations.


