Question

What is the equation of the parabola with focus (1,1/2) and directrix y=3

Answer 1

In this problem, we have a vertical parabola open downward

The vertex is a maximum

Remember that

The distance from the vertex to the directrix must be the same that the distance from the vertex to the focus

so

step 1

Find out the distance from the directrix to the focus

directrix -------> y=3

Focus ----> (1,1/2)

distance=3-1/2=2.5

2.5/2=1.25

the vertex is the midpoint between the focus and the directrix

so

the coordinates of the vertex are (1,0.5+1.25)

vertex (1,1.75)--------> vertex (1.7/4)

The equation in vertex form is given by

`y=a(x-h)^2+k`

where

a is the leading coefficient

(h,k) is the vertex

(h,k)=(1,7/4)

substitute

`y=a(x-1)^2+(7)/(4)`
`y=a(x^2-2x+1)+(7)/(4)`

Remember that

`-(x-h)^2=4p(y-k)`

where

p is the focal distance

In this problem

p=1.25=5/4

substitute (h,k)=(1,7/4)

`-(x-1)^2=4((5)/(4))(y-(7)/(4))`
`-x^2+2x-1=5y-(35)/(4)`

Isolate the variable y

`\begin{gathered} 5y=-x^2+2x-1+(35)/(4) \\ 5y=-x^2+2x+(31)/(4) \\ \\ y=-(1)/(5)x^2+(2)/(5)x+(31)/(20) \end{gathered}`

The answer is option C