Question

Derive the equation of the parabola with a focus at (−5, 5) and a directrix of y = −1. (2 points)

f(x) = − one twelfth (x − 5)2 + 2

f(x) = one twelfth (x − 5)2 + 2

f(x) = − one twelfth (x + 5)2 + 2

f(x) = one twelfth (x + 5)2 + 2

Answer 1

`f(x)=(1)/(12)(x + 5) ^ 2 +2`

Explanation:

With the focus and the directrix we can find the equation of the parabola

Imagine any point belonging to the parabola, let's call that point (x, y).

If the point (x, y) belongs to the parable sought then:

The distance between the point and the focus is:

`√((x -(-5)) ^ 2 + (y-5) ^ 2)`

The distance between the point and the directrix is:

|y -(-1)|

These distances are the same for any point belonging to the parabola. Then we equal them:

`√((x+5) ^ 2 + (y-5) ^ 2)= |y+1|`

We raise both sides of the equation squared and we have left:

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

Grouping equal terms and simplifying:

`(x + 5) ^ 2 = [y ^ 2 + 2y +1] - [y ^ 2 -10y +25]\\ (x + 5) ^ 2 = 12y-24\\ 12y = (x + 5) ^ 2 +24\\ y = (1)/(12) (x + 5) ^ 2 +2`

As this condition is fulfilled for all the points in the parabola, then:

`f(x)=(1)/(12)(x + 5) ^ 2 +2` It is the equation of the parable sought, and its derivative is:

`(df(x))/(dx)=(1)/(6)(x + 5)`

Answer 2

Focus of the parabola is (-5,5) and directrix is y = -1.

Let's assume a point (x,y) on parabola.

According to definition of parabola, the distance between point (x,y) and focus (-5,5) would be same as the distance between the point (x,y) and directrix y = -1.

`√((x+5)^2+(y-5)^2) = √((y+1)^2) \\\\(x+5)^2+(y-5)^2=(y+1)^2 \\\\(x+5)^2 + y^2 - 10y +25 = y^2 +2y +1 \\\\(x+5)^2 =  -y^2 +10y -25 + y^2 +2y +1 \\\\(x+5)^2 = 12y -24 \\\\12y =  (x+5)^2  +24 \\\\ y = (1)/(12) (x+5)^2  +2`

Hence, option D is correct, i.e. f(x) = one twelfth (x + 5)2 + 2.