Question

What is the equation in standard form of the parabola with a focus at (42,-31) and a directrix of y= 2?

Answer 1

`y = -1/66x^2 +14/11x- 907/22 larr` standard form

Explanation:

Please observe that the directrix is a horizontal line

`y = 2`

Therefore, the parabola is the type that opens upward or downward; the vertex form of the equation for this type is:

`y = 1/(4f)(x -h)^2+ k" [1]"`

Where `(h,k)` is the vertex and `f` is the signed vertical distance from the vertex to the focus.

The x coordinate of the vertex is the same as the x coordinate of the focus:

`h = 42`

Substitute `42` for `h` into equation [1]:

`y = 1/(4f)(x -42)^2+ k" [2]"`

The y coordinate of the vertex is halfway between the directrix and the focus:

`k = (y_"directrix"+y_"focus")/2`

`k = (2+(-31))/2`

`k = -29/2`

Substitute `-29/2` for `k` into equation [2]:

`y = 1/(4f)(x -42)^2-29/2" [3]"`

The equation to find the value of `f` is:

`f = y_"focus"-k`

`f = -31- (-29/2)`

`f = -33/2`

Substitute `-33/2` for `f` into equation [3]:

`y = 1/(4(-33/2))(x -42)^2-29/2`

Simplify the fraction:

`y = -1/66(x -42)^2-29/2`

Expand the square:

`y = -1/66(x^2 -84x+ 1764)-29/2`

Distribute the fraction:

`y = -1/66x^2 +14/11x- 294/11-29/2`

Combine like terms:

`y = -1/66x^2 +14/11x- 907/22 larr` standard form

Answer 2

`y=-1/66x^2+14/11x-907/22,`

Explanation:

We will solve this Problem using the following Focus-Directrix

Property (FDP) of the Parabola.

FDP : Any point on a Parabola is equidistant from the

Focus and the Directrix.

Let, the point `F=F(42,-31)," and, the line "d : y-2=0,` be

the Focus and the Directrix of the Parabola, say S.

Let, `P=P(x,y) in S,` be any General Point.

Then, using the Distance Formula, we have, the distance,

`FP=sqrt{(x-42)^2+(y+31)^2}...............................(1).`

Knowing that the `bot-`dist. between a point `(k,k),` and, a line :

`ax+by+c=0,` is, `|ah+bk+c|/sqrt(a^2+b^2),` we find that,

`"the "bot-"dist. btwn "P(x,y), &, d" is, "|y-2|..............(2).`

By FDP, `(1), and (2),` we have,

`sqrt{(x-42)^2+(y+31)^2}=|y-2|, or,`

`(x-42)^2=(y-2)^2-(y+31)^2=-66y-957, i.e.,`

`x^2-84x+1764=-66y-957.`

`:. 66y=-x^2+84x-2721,` which, in the Standard Form,

reads, `y=-1/66x^2+14/11x-907/22,`

as Respected Douglas K. Sir has already derived!

Enjoy Maths.!