Let $X$ be an irreducible hypersurface defined by a polynomial $f$ of degree $5$ in $\mathbb{P}^3$. Let the homogeneous co-ordinates is given by $[x, y, z, w]$ and let $H$ be a hyperplane given by $w= 0$. Assume that the intersection $X \cap H$, a curve of degree $5$ in the plane $H$, is given by an equation of the form $hg^2=0$, where $h$ is a linear form and $g$ is of degree $2$. Further assume that $X$ is singular along the plane curve $g= 0$. In this situation one can write $f$ as $f = hg^2 + wgf\_2 + w^2f\_3$, where $f\_2$ is a polynomial of degree $2$ in $x, y, z$. My question is the following:
Can $X$ have isolated singularities? If yes then is it possible to give an upper bound on the number of isolated singularity ?


Thanks in advance.



singularity of a hypersuface in $\mathbb{P}^3$
Consider the blowup $BL\_C\mathbb{P}^3$ of $\mathbb{P}^3$ along the conic 
$$
C = \{ w = g = 0 \} \subset \mathbb{P}^3.
$$
Let $H$ be the pullback of the class of a hyperplane in $\mathbb{P}^3$ and $E \subset Bl\_C\mathbb{P}^3$ is the exceptional divisor. Then the linear system of quintics in $\mathbb{P}^3$ singular along $C$ identifies with the linear system $|5H - 2E|$ on the blowup. Furthermore, a general divisor in $|5H - 2E|$ is the blowup of such a quintic along $C$, i.e., its normalization. Thus, your question is about Betti numbers of a general divisor in $Bl\_C\mathbb{P}^3$ from the linear system $|5H - 2E|$.


Next, note that
$$
Bl\_C\mathbb{P}^3 \cong Bl\_PQ^3,
$$
the blowup of a 3-dimensional quadric at a point. Furthermore, if $H'$ and $E'$ are the preimage of the hyperplane class of $Q^3$, and the exceptional divisor of the econd vblowup, then
$$
H = H' - E',\qquad E = H' - 2E'.
$$
Therefore,
$$
5H - 2E = 5(H' - E') - 2(H' - 2E') = 3H' - E',
$$
hence the required surface is the strict transform of a general cubic hypersurface through $P$. This means it is isomorphic to the blowup at point of $Y$, the intersection of a quadric and cubic in $\mathbb{P}^4$. The Betti numbers of $Y$ are well known, these are
$$
b\_0(Y) = b\_2(Y) = b\_4(Y) = b\_6(Y) = 1,\qquad b\_3(Y) = 40,
$$
hence the Betti numbers are
$$
b\_0 = b\_6 = 1,\qquad b\_2 = b\_4 = 2, \qquad b\_3 = 40.
$$


