Is the space of real conics with a singular point an orientable manifold? 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.  
 A: 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.
