Question

What is the equation of the parabola with a focus at (44,55) and a directrix of y= 66?

Answer 1

`x^2-88x+22y+605=0`

Explanation:

Parabola is the locus of a point which moves so that its distances from a given point called focus and from a given line called directrix are equal.

Here let us consider the point as `(x,y)`. Its distance from focus `(44,55)` is `sqrt((x-44)^2+(y-55)^2)`

and as distance of a point `x_1,y_1)` from a line `ax+by+c=0` is `|(ax_1+by_1+c)/sqrt(a^2+b^2)|`, distance of `(x,y)` from `y=66` or `y-66=0` (i.e. `a=0` and `b=1`) is `|y-66|`.

Hence equation of parabola is

`(x-44)^2+(y-55)^2=(y-66)^2`

or `x^2-88x+1936+y^2-110y+3025=y^2-132y+4356`

or `x^2-88x+22y+605=0`

The parabola along with focus and directrix appears as shown below.

graph{(x^2-88x+22y+605)((x-44)^2+(y-55)^2-6)(y-66)=0 [-118, 202, -82.6, 77.4]}

Answer 2

`y=-1/18(x^2-88x+847)`

Explanation:

Focus `(44, 55)`
Directrix `y=66`

Vertex `(44, (55+66)/2)=(44,60.5)`

Distance between vertex and focus `a= 60.5-55=4.5`

Since Directrix is above vertex, this parabola opens down.
Its equation is -

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

Where -

`h=44`
`k=60.5`
`a=4.5`

`(x-44)^2=-4xx4.5(y-60.5)`
`x^2-88x+1936=-18y+1089`

`-18y+1089=x^2-88x+1936`

`-18y=x^2-88x+1936-1089`
`-18y=x^2-88x+847`

`y=-1/18(x^2-88x+847)`