The problem asks us to find the coordinates of the focus and the equation of the directrix for each of the given parabolas. We will address problems 1 through 8.

1. Problem Description

The problem asks us to find the coordinates of the focus and the equation of the directrix for each of the given parabolas. We will address problems 1 through

8.

2. Solution Steps

Problem 1: $y^2 = 4x$

The general form is $y^2 = 4px$. Comparing, $4p = 4$, so $p = 1$.

The focus is at $(p, 0)$, which is $(1, 0)$.

The directrix is $x = -p$, which is $x = -1$.

Problem 2: $y^2 = -12x$

The general form is $y^2 = 4px$. Comparing, $4p = -12$, so $p = -3$.

The focus is at $(p, 0)$, which is $(-3, 0)$.

The directrix is $x = -p$, which is $x = 3$.

Problem 3: $x^2 = -12y$

The general form is $x^2 = 4py$. Comparing, $4p = -12$, so $p = -3$.

The focus is at $(0, p)$, which is $(0, -3)$.

The directrix is $y = -p$, which is $y = 3$.

Problem 4: $x^2 = -16y$

The general form is $x^2 = 4py$. Comparing, $4p = -16$, so $p = -4$.

The focus is at $(0, p)$, which is $(0, -4)$.

The directrix is $y = -p$, which is $y = 4$.

Problem 5: $y^2 = x$

The general form is $y^2 = 4px$. Comparing, $4p = 1$, so $p = \frac{1}{4}$.

The focus is at $(p, 0)$, which is $(\frac{1}{4}, 0)$.

The directrix is $x = -p$, which is $x = -\frac{1}{4}$.

Problem 6: $y^2 + 3x = 0$

Rewrite the equation as $y^2 = -3x$.

The general form is $y^2 = 4px$. Comparing, $4p = -3$, so $p = -\frac{3}{4}$.

The focus is at $(p, 0)$, which is $(-\frac{3}{4}, 0)$.

The directrix is $x = -p$, which is $x = \frac{3}{4}$.

Problem 7: $6y - 2x^2 = 0$

Rewrite the equation as $2x^2 = 6y$, or $x^2 = 3y$.

The general form is $x^2 = 4py$. Comparing, $4p = 3$, so $p = \frac{3}{4}$.

The focus is at $(0, p)$, which is $(0, \frac{3}{4})$.

The directrix is $y = -p$, which is $y = -\frac{3}{4}$.

Problem 8: $3x^2 - 9y = 0$

Rewrite the equation as $3x^2 = 9y$, or $x^2 = 3y$.

The general form is $x^2 = 4py$. Comparing, $4p = 3$, so $p = \frac{3}{4}$.

The focus is at $(0, p)$, which is $(0, \frac{3}{4})$.

The directrix is $y = -p$, which is $y = -\frac{3}{4}$.

3. Final Answer

1. Focus: $(1, 0)$, Directrix: $x = -1$

2. Focus: $(-3, 0)$, Directrix: $x = 3$

3. Focus: $(0, -3)$, Directrix: $y = 3$

4. Focus: $(0, -4)$, Directrix: $y = 4$

5. Focus: $(\frac{1}{4}, 0)$, Directrix: $x = -\frac{1}{4}$

6. Focus: $(-\frac{3}{4}, 0)$, Directrix: $x = \frac{3}{4}$

7. Focus: $(0, \frac{3}{4})$, Directrix: $y = -\frac{3}{4}$

8. Focus: $(0, \frac{3}{4})$, Directrix: $y = -\frac{3}{4}$