Question

: This parabola opens to the right.

X^2 = 12y

A: True
B: False


: Write the coordinate point for the vertex of this parabola:

1
x = - ---- y^2
8
Answer:_____

: What is the value of p? x^2 = 12y_______

: What is the length of the focal width? X^2 = 12y______

Answer 1

Answer on Question #57312 – Math – Analytic Geometry

Question

1. This parabola opens to the right.

$$x^2 = 12y$$

A: True

B: False

Solution

In this question the $x$ is squared, so this parabola is vertical, the leading coefficient is positive $y = \frac{1}{12} x^2$, $\frac{1}{12} > 0$, therefore it opens upwards (like U), it does not open to the right.

Answer: B: False.

Question

2. Write the coordinate point for the vertex of this parabola:

$$x = -\frac{1}{8} y^2.$$

Solution

The conics form of the sideways parabola equation is $4p(x - h) = (y - k)^2$ and the vertex is $(h, k)$. We have

$(x - 0) = -\frac{1}{8} (y - 0)^2$ or $-8(x - 0) = (y - 0)^2$, the vertex is $(h, k) = (0, 0)$.

Answer: $(0, 0)$.

Question

3. What is the value of $p$? $x^2 = 12y$.

Solution

The conics form of the regular parabola equation is $4p(y - k) = (x - h)^2$

We have $4p(y - 0) = (x - 0)^2$, $4p = 12$, $p = 3$.

Answer: $p = 3$.

Question

4. What is the length of the focal width? $x^2 = 12y$.

Solution

The value $|4p|$ is the focal width of the parabola. From Part 3 we know $p = 3$, therefore, the length of the focal width is $4 \cdot 3 = 12$.

Answer: 12.

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