Let’s find the equations of the parabolas for each part step by step:
### Part (a)
Given:
- Vertex [tex]\( V(0, 0) \)[/tex]
- Focus [tex]\( F(0, 4) \)[/tex]
A parabola with its vertex at the origin [tex]\( (0, 0) \)[/tex] and focus [tex]\((0, 4)\)[/tex] opens upwards. The standard form of a parabola opening upwards is [tex]\(x^2 = 4py\)[/tex], where [tex]\( p \)[/tex] is the distance from the vertex to the focus.
Since the distance [tex]\( p \)[/tex] is 4, substituting this value into the equation gives:
[tex]\[ x^2 = 4 \cdot 4 \cdot y \][/tex]
So, the equation of the parabola is:
[tex]\[ x^2 = 16y \][/tex]
### Part (b)
Given:
- Vertex [tex]\( V(0, 0) \)[/tex]
- Directrix [tex]\( x = -5 \)[/tex]
A parabola with its vertex at the origin [tex]\((0, 0)\)[/tex] and directrix [tex]\( x = -5 \)[/tex] opens to the right. The standard form for a parabola opening to the right is [tex]\( y^2 = 4px \)[/tex], where [tex]\( p \)[/tex] is the distance from the vertex to the directrix.
The distance [tex]\( p \)[/tex] is 5. Substituting into the equation gives:
[tex]\[ y^2 = 4 \cdot 5 \cdot x \][/tex]
So, the equation of the parabola is:
[tex]\[ y^2 = 20x \][/tex]
### Part (c)
Given:
- Vertex [tex]\( V(1, 4) \)[/tex]
- Focus [tex]\( F(-2, 4) \)[/tex]
A parabola that is horizontally oriented with vertex [tex]\( (1, 4) \)[/tex] and focus [tex]\( (-2, 4) \)[/tex] opens to the left. The standard equation for a horizontally oriented parabola is [tex]\( (y - k)^2 = 4p(x - h) \)[/tex], where [tex]\( (h, k) \)[/tex] is the vertex, and [tex]\( p \)[/tex] is the distance from the vertex to the focus. Here, the distance [tex]\( p \)[/tex] between the vertex and the focus is [tex]\( 1 - (-2) = 3 \)[/tex].
So, the equation for the parabola is:
[tex]\[ (y - 4)^2 = 4 \cdot 3 \cdot (x - 1) \][/tex]
Simplifying gives:
[tex]\[ (y - 4)^2 = 12(x - 1) \][/tex]
### Part (d)
Given:
- Focus [tex]\( F(2, 3) \)[/tex]
- Directrix [tex]\( y = -1 \)[/tex]
For a parabola that opens upwards or downwards, and given a focus [tex]\( (h, k + p) \)[/tex] and a directrix [tex]\( y = k - p \)[/tex]:
The midpoint between the directrix and the focus gives the vertex. Here, using the focus [tex]\( F(2, 3) \)[/tex] and directrix [tex]\( y = -1 \)[/tex], we find:
[tex]\[ p = \frac{3 - (-1)}{2} = 2 \][/tex]
[tex]\[ \text{Vertex: } (2, 3-2) = (2, 1) \][/tex]
The standard equation for a parabola opening upwards is [tex]\( (x - h)^2 = 4p(y - k) \)[/tex].
Therefore, the equation for the parabola is:
[tex]\[ (x - 2)^2 = 4 \cdot 2 \cdot (y - 1) \][/tex]
Simplifying, we get:
[tex]\[ (x - 2)^2 = 8(y - 1) \][/tex]
### Part (e)
Given:
- Vertex [tex]\( V(0, 0) \)[/tex]
- Passing through [tex]\( (4, 5) \)[/tex]
A parabola with vertex at the origin [tex]\( (0, 0) \)[/tex] and given to be passing through [tex]\( (4, 5) \)[/tex] and horizontal axis [tex]\( y = 0 \)[/tex] will have the standard form of a parabola opening upwards, which is:
[tex]\[ y = ax^2 \][/tex]
Given the point [tex]\((4, 5)\)[/tex], substitute [tex]\( x = 4 \)[/tex] and [tex]\( y = 5 \)[/tex]:
[tex]\[ 5 = a \cdot (4^2) = 16a \][/tex]
[tex]\[ a = \frac{5}{16} \][/tex]
So, the equation:
[tex]\[ y = \frac{5}{16}x^2 \][/tex]
Therefore, the equation is:
[tex]\[ y = 0.3125x^2 \][/tex]
