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professor190 [17]
3 years ago
7

Write an equation of a parabola with a vertex at the origin and a directx at y=5

Mathematics
1 answer:
VMariaS [17]3 years ago
4 0
Directrix at y=5 (horizontal line) → The parabola is vertical
Vertex at the orgin: V=(0,0)=(h,k)→h=0, k=0
The directix is above the vertex→the parabola opens downward→p is negative
p is the distance between the vertex and the directrix→p=-5
Equation has the form:
(x-h)^2=4p(y-k)
(x-0)^2=4(-5)(y-0)
x^2=-20y
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Consider parallelogram VWXY below.
olchik [2.2K]

note:Parallelogram WXYZ not VWXY & I will answer the questions in the image b'cos I don't get your question.

Answers:

a.(i.)Slope of line YZ = 4/5

(ii.)Line adjacent line YZ is line YX,slope of YX =5/4

b.(i.)Length of line YZ =√41 units

 (ii.)Length of line YX =√41 units

c.WXYZ is a rhombus.

Step-by-step explanation:

a.Slope=gradient(m)=(y₂-y₁)/(x₂-x₁)

(i.)m of line YZ=(y₂-y₁)/(x₂-x₁) where Y(-6,-1) & Z(-1,3)

(I believe you know what the y₂,y₁,x₂ & x₁ stand for.)

So m=(y₂-y₁)/(x₂-x₁),=(3-(-1))/(-1-(-6) =(4)/(5)=4/5

∴m of line YZ is 4/5

(ii.)m of line YX=(y₂-y₁)/(x₂-x₁) where Y(-6,-1) & X(-2,4)

So m=(y₂-y₁)/(x₂-x₁),=(4-(-1))/(-2-(-6) =(5)/(4)=5/4=1.25

∴m of line YX is 5/4

b.(i.)To find the length of line YZ,we have to express it as column vector YZ.

Vector YZ=OZ-OY,Z-Y=(-1,3)-(-6,-1)=(5,4)

Length=√x² plus y²=√5² plus 4²=√25 plus 16=√41 units

∴The length of line YZ is √41 units.

(ii.)To find the length of line YX,we have to express it as column vector YZ.

Vector YX=OX-OY,X-Y=(-2,4)-(-6,-1)=(4,5)

Length=√x² plus y²=√4² plus 5²=√16 plus 25=√41 units

∴The length of line YX is √41 units.

c.WXYZ is a rhombus.

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2 years ago
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