Q.Find the equation of the parabola in each of the cases given below:

Comparing $x^2=24y$ with the standard form $x^2=4ay$ directly gives the latus rectum length as $24$.

1. The standard form of a parabola opening upward is $x^2=4ay$, whose latus rectum has length $4a$.
2. Comparing the given equation $x^2=24y$ with $x^2=4ay$, we get $4a=24$.
3. The length of the latus rectum is defined as exactly $4a$, so it equals $24$ directly (no need to solve for $a$ separately).
4. (Equivalently, $a=6$, and $4a=4(6)=24$, confirming the same result.)

Rewriting $x^2=8y-1$ in the standard form $x^2=4a(y-k)$ shows the vertex is $\left(0,\dfrac{1}{8}\right)$.

1. Start with $x^2=8y-1$.
2. Add $1$ to both sides and factor the right side: $x^2 = 8y-1 = 8\left(y-\dfrac{1}{8}\right)$.
3. This is now in the standard vertical-axis parabola form $x^2=4a(y-k)$, where the vertex is $(h,k)=(0,k)$ (here $h=0$ since there is no $x$-shift, only a coefficient of $x^2$).
4. Comparing $8\left(y-\dfrac{1}{8}\right)$ with $4a(y-k)$, we get $k=\dfrac{1}{8}$ (and $4a=8$, i.e. $a=2$).
5. So the vertex is at $\left(0,\dfrac{1}{8}\right)$.

The length of the latus rectum is $4a=8$, where $a$ is the vertex-to-directrix distance.

1. For a parabola, the vertex lies midway between the focus and the directrix, at distance $a$ from each.
2. Vertex $=(2,-3)$, directrix $x=4$ (a vertical line), so the axis of the parabola is horizontal, $y=-3$.
3. Distance from vertex to directrix: $a=|4-2|=2$.
4. Length of latus rectum $=4a=4(2)=8$.
5. Options (a) $2$, (b) $4$, (c) $6$ correspond to using $a$, $2a$, or another wrong multiple instead of $4a$.