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A parabola can be drawn given a focus of (7, -5)(7,−5) and a directrix of x=3x=3. What can be said about the parabola?
The parabola has a vertex at ( , ), has a p-value of
and it
-opens up
-opens down
-opens right
-opens left

A Parabola Can Be Drawn Given A Focus Of 7 575 And A Directrix Of X3x3 What Can Be Said About The Parabola The Parabola Has A Vertex At Has A Pvalue Of And It O class=

Sagot :

Step-by-step explanation:

Since the focus is on the right in relation to the directrix, the parabola will open right.

Next, for a parabola opening to the right. the formula is

[tex](y - k) {}^{2} = 4p(x - h)[/tex]

where (h,k) is the vertex.

P is the midpoint of the total distance between the focus and directrix.

Since the parabola is opening right, our vertex and focus will lie on the x axis.

The vertex lies halfway between directrix and focus so the vertex is at

(5,-5).

Note: We choose the point (3,-5) for the directrix because the focus also have (7,-5).

This means p=2.

[tex](y + 5) {}^{2} = 4(2)(x - 5)[/tex]

So our vertex is (5,-5) p=2,

opens right

VectorFundament120idn

Answer:

Vertex = (5,-5)

P-value = 2

Opens right

Step-by-step explanation:

Given:

  • Focus = (7,-5)
  • Directrix = x = 3

Since focus is on the right side of directrix, it obviously opens right.

Focus for right/left parabola is defined as (h+p,k) and directrix is defined as x = h - p, we’ll be using simultaneous equation for both directrix and focus equation to find vertex and p-value.

[tex]\displaystyle \large{h+p = 7 \to (1)}\\\displaystyle \large{h-p = 3 \to (2)}[/tex]

Solve the simultaneous equation:

[tex]\displaystyle \large{2h=10}\\\displaystyle \large{h=5}[/tex]

Substitute h = 5 in any equation but I’ll choose (1).

[tex]\displaystyle \large{5+p=7}\\\displaystyle \large{p=2}[/tex]

Therefore, from (h,k), the vertex is at (5,-5) and with p-value of 2, since p > 0 then the parabola opens right.

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