88.7k views
5 votes
G find the standard form of the equation of the parabola satisfying the given conditions. ​focus: left parenthesis negative 5 comma 6 right parenthesis​; ​directrix: y equals 2

User Rey
by
8.3k points

2 Answers

4 votes

Answer:

dee um quadratic graphs and their properties quick check tingz

1. C ,->(1,-4)

2. B -> (y=0.5^2+1)

3. D -> (the graph that goes like down and has -4)

4. A -> (f(x) = –10x², f(x) = 2x², f(x) = 0.5x²)

Explanation:

im valid bc i did the test <333

User Pierre Vieira
by
7.9k points
4 votes

Answer:

y = 1/8x² +5/4x +57/8

Explanation:

You want the standard form of the equation for the parabola with focus (-5, 6) and directrix y = 2.

Vertex

The vertex is halfway between the focus with coordinate y=6 and the directrix with equation y=2. Its y-coordinate will be (6+2)/2 = 4. It is 2 units below the focus, so p=2. The x-coordinate of the vertex is the same as that of the focus, so the vertex is (-5, 4).

Equation

The vertex form equation for the parabola is ...

y = 1/(4p)(x -h)² +k . . . . . . . vertex at (h, k)

y = 1/8(x +5)² +4 . . . . . . standard form in UK

Expanding this equation, we get ...

y = 1/8x² +5/4x +57/8 . . . . . . standard form in US

__

Additional comment

The "standard form" of the equation for a parabola varies by location.

<95141404393>

G find the standard form of the equation of the parabola satisfying the given conditions-example-1
User Predominant
by
8.2k points