Since you know the explicit equation of the conic, you can compute everything by using Macaulay2.

The following script should be clear:

```
i1 : k=ZZ/32003;
i2 : ringP1=k[x, y];
i3 : ringP4=k[z1, z2, z3, z4, z5];
i4 : I= ideal(z1^5+z2^5+z3^5+z4^5+z5^5-z1*z2*z3*z4*z5);
o4 : Ideal of ringP4
i5 : ringQuintic=ringP4/I;
i6 : conicMap=map(ringP1, ringQuintic, {x^2, -x^2, y^2, -y^2, x*y});
o6 : RingMap ringP1 <--- ringQuintic
i7 : conic=image conicMap;
i8 : IC=ideal conic;
o8 : Ideal of ringQuintic
i9 : ConormalModuleConic = IC/IC^2;
i10 : ConormalSheafConic= sheaf ConormalModuleConic;
i11 : NormalSheafConic= dual sheaf ConormalModuleConic;
i12 : HH^0(ConormalSheafConic)
4
o12 = k
o12 : k-module, free
i13 : HH^1(ConormalSheafConic)
o13 = 0
o13 : k-module
i14 : HH^0(NormalSheafConic)
o14 = 0
o14 : k-module
i15 : HH^1(NormalSheafConic)
o15 = 0
o15 : k-module
```

The output reads

$h^0(X, N_{C|X}^{*})=4, \quad h^1(X, N_{C|X}^{*})=0, \quad h^0(X, N_{C|X})=0, \quad h^1(X, N_{C|X})=0$,

hence $N_{C|X}=\mathcal{O}_{\mathbb{P}^1}(-1) \oplus \mathcal{O}_{\mathbb{P}^1}(-1)$, according to Sasha's and Sandor's answers.