Question

HELLLPPP!!!

What is the equation of the parabola with focus(1, 1/2)
and directrix y = 3?
A. Y=x^2+17x-1
B. Y=-3/16x^2
C. Y=1/5x^2+2/5x+31/20
D. Y=-3/5x^2-14/31x+71/16

Answer 1

Given:

Focus of a parabola =
`\left(1,(1)/(2)\right)`

Directrix:
`y=3`

To find:

The equation of the parabola.

Solution:

The equation of a vertical parabola is:

`y-k=(1)/(4a)(x-h)^2` ...(A)

Where,
`(h,k)` is center,
`(h,k+a)` is focus and
`y=k-a` is the directrix.

On comparing the focus, we get

`(h,k+a)=\left(1,(1)/(2)\right)`

`h=1`

`k+a=(1)/(2)` ...(i)

On comparing the directrix, we get

`k-a=3` ...(ii)

Adding (i) and (ii), we get

`2k=(7)/(2)`

`k=(7)/(4)`

Putting
`k=(7)/(4)` is (i), we get

`(7)/(4)+a=(1)/(2)`

`a=(1)/(2)-(7)/(4)`

`a=(-5)/(4)`

Putting
`a=(-5)/(4),h=1,k=(7)/(4)` in (A), we get

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

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

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

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

On further simplification, we get

`y=-(1)/(5)x^2+(2)/(5)x+(35-4)/(20)`

`y=-(1)/(5)x^2+(2)/(5)x+(31)/(20)`

Therefore, the equation of the parabola is
`y=-(1)/(5)x^2+(2)/(5)x+(31)/(20)`.

Note: Option C is correct but the leading coefficient should be negative.