Given trajectory

The particle moves such that $y = \frac{x^2}{2}.$

Since motion is in the plane, let $x = x(t), \qquad y = y(t)=\frac{x(t)^2}{2}.$

Velocity components

Differentiate with respect to time: $\dot y = x\dot x.$ Hence, $v_x = \dot x, \qquad v_y = \dot y = x\dot x.$

So the velocity vector is $\vec v = (\dot x,\, x\dot x).$

Use the initial speed condition

At $t=0$, the particle starts from the origin, so $x(0)=0, \qquad y(0)=0.$

Initial speed is $1\,\text{m s}^{-1}$, so $|\vec v(0)| = 1.$

But at $x=0$, $v_y(0)=x(0)\dot x(0)=0.$ Therefore, $|\vec v(0)| = |v_x(0)| = |\dot x(0)| = 1.$

So initially the velocity is entirely along the x-axis (either $+x$ or $-x$ direction).

Hence Option C is correct.

Acceleration components

Differentiate $\dot y = x\dot x$ again: $\ddot y = \dot x^2 + x\ddot x.$

Therefore, $a_x = \ddot x, \qquad a_y = \dot x^2 + x\ddot x.$

So $a_y = v_x^2 + x a_x.$

Check Option A

Option A states: if $a_x=1\,\text{m s}^{-2}$, then when the particle is at the origin, $a_y=1\,\text{m s}^{-2}$.

At the origin, $x=0$, so $a_y = \dot x^2 + x\ddot x = \dot x^2.$

At $t=0$ (starting from origin), we already found $|\dot x(0)|=1 \implies \dot x(0)^2=1.$

Thus, $a_y(0)=1\,\text{m s}^{-2}.$

This is true irrespective of the sign of $\dot x(0)$, and the given condition $a_x=1$ is consistent.

Hence Option A is correct.

Check Option B

If $a_x=0$, then $\ddot x=0 \implies \dot x=\text{constant}.$

Since initial speed at the origin is $1$ and initially velocity is along x-axis, $|\dot x|=1.$ Therefore, $\dot x^2=1 \quad \text{for all times}.$

Now using $a_y = \dot x^2 + x a_x,$ with $a_x=0$, $a_y = \dot x^2 = 1.$

So at all times, $a_y=1\,\text{m s}^{-2}.$

Hence Option B is correct.

Check Option D

Again, if $a_x=0$, then $\dot x$ is constant with magnitude $1$.

Taking the usual positive-x start, $\dot x=1$, so $x=t$ because $x(0)=0$.

Then $v_y = x\dot x = t\cdot 1 = t.$

At $t=1\,\text{s}$, $v_x=1, \qquad v_y=1.$

Therefore, $\tan\theta = \frac{v_y}{v_x} = 1 \implies \theta=45^\circ.$

Hence Option D is correct.

Final conclusion

All four statements are correct: $\boxed{A, B, C, D}$