Conic Sections - Standard equations of parabola

Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum for the equation $y^2 = 12x$.

Comparing the given equation $y^2 = 12x$ with the standard form $y^2 = 4ax$, we get: $4a = 12$ $a = \frac{12}{4} = 3$

1. Focus: The focus is $(a, 0)$, which is $(3, 0)$.
2. Axis: The axis of the parabola is the $x$-axis, so the equation is $y = 0$.
3. Directrix: The equation of the directrix is $x = -a$, which is $x = -3$.
4. Length of Latus Rectum: $4a = 12$.

Since the equation is of the form $y^2 = 4ax$, the parabola opens to the right along the $x$-axis.

Find the equation of the parabola that satisfies the following conditions: Focus $(0, -3)$ and Directrix $y = 3$.

The focus is $(0, -3)$, which is of the form $(0, -a)$. This implies $a = 3$. Since the focus lies on the $y$-axis and is below the origin, the parabola opens downwards. The standard equation for a parabola opening downwards is $x^2 = -4ay$. Substituting $a = 3$: $x^2 = -4(3)y$ $x^2 = -12y$

The focus $(0, -3)$ tells us the axis of symmetry is the $y$-axis and the vertex is at the origin $(0, 0)$ because the vertex is the midpoint between the focus and the directrix. Since the focus has a negative $y$-coordinate, the parabola opens downwards.