Answer:
Four parabolas are possible.
Step-by-step explanation:
Vertex is at (0,0)
So it is a standard parabola.
Focus can be at four possible points: (5,0),(0,5),(-5,0),(0,-5)
So, a=5
Four possible parabolas are
Upward Parabola (x²=4ay)
Downward Parabola (x²=-4ay)
Rightward Parabola (y²=4ax)
Leftward Parabola (y²=-4ax)
So if you put value of "a" in above four equations you get
x²=20y
x²=-20y
y²=20x
y²=-20x
Answer:
4 possible parabolas:
Step-by-step explanation:
In standard Cartesian coordinates, parabolas can be either vertical or horizontal. A vertical parabola can open upward or downward, and a horizontal parabola can open left or right. Therefore, if a parabola has its vertex at the origin (0, 0) and the distance from the vertex to the focus is 5 units, there are four possible parabolas that fit this description.
[tex]\dotfill[/tex]
The standard form of a vertical parabola is:
[tex](x-h)^2=4p(y-k)[/tex]
where (h, k) is the vertex, and p is the distance from the vertex to the focus.
Given that the vertex is at the origin (0, 0) and the distance from the vertex to the focus is 5 units, then:
Substitute these values into the formula to create two equations:
[tex](x-0)^2=4(-5)(y-0) \implies x^2=-20y\\\\(x-0)^2=4(5)(y-0)\implies x^2=20y[/tex]
[tex]\dotfill[/tex]
The standard form of a horizontal parabola is:
[tex](y-k)^2=4p(x-h)[/tex]
where (h, k) is the vertex, and p is the distance from the vertex to the focus.
Given that the vertex is at the origin (0, 0) and the distance from the vertex to the focus is 5 units, then:
Substitute these values into the formula to create two equations:
[tex](y-0)^2=4(-5)(x-0)\implies y^2=-20x\\\\(y-0)^2=4(5)(x-0)\implies y^2=20x\\\\[/tex]
[tex]\dotfill[/tex]
Therefore, the equations of all possible parabolas are:
[tex]x^2 = -20y\\\\x^2=20y\\\\y^2=-20x\\\\y^2=20x[/tex]
