Question

Derive the equation of the parabola with a focus at (6, 2) and a directrix of y = 1. f(x) = −one half(x − 6)2 + three halves f(x) = one half(x − 6)2 + three halves f(x) = −one half(x + three halves)2 + 6 f(x) = one half(x + three halves)2 + 6

Answer 1

Answer:

Second choice.

f(x) = 1/2(x - 6)^2 + 3/2.

Step-by-step explanation:

The distance of a point  (x, y) from the focus = the distance of the point from the directrix, so:

(x - 6)^2 + (y - 2)^2 = (y - 1)^2

x^2 - 12x + 36 + y^2 - 4y + 4 = y^2 - 2y + 1

x^2 -12x + 39 = 2y

y = f(x) = 1/2 (x^2 - 12x + 39)

I see you want the answer in vertex for so it is:

f(x)  = 1/2 [ (x - 6)^2 - 36)  + 39)

f(x)  = 1/2(x - 6)^2 + 3)

f(x) = 1/2(x - 6)^2 + 3/2.

Answer 2

A parabola is a plane that is approximately U-shaped.

The equation of the parabola is: $\mathbf{y = \frac{1}{2}(x - 6)^2 + \frac 32}$

The given parameters are:

$\mathbf{Focus = (6,2)}$

$\mathbf{Directrix: y = 1}$

First, equate the directrix to 0

$\mathbf{y - 1 = 0}$

The equation is then calculated as:

$\mathbf{(x - a)^2 + (y - b)^2 = (y- 1)^2}$

Where:

$\mathbf{(a,b) = (6,2)}$

So, we have:

$\mathbf{(x - 6)^2 + (y - 2)^2 = (y- 1)^2}$

Expand

$\mathbf{x^2 - 12x +36 + y^2 - 4y + 4 = y^2 - 2y + 1}$

Subtract y^2 from both sides

$\mathbf{x^2 - 12x +36 - 4y + 4 =- 2y + 1}$

Collect like terms

$\mathbf{x^2 - 12x +36 + 4 - 1 =4y - 2y}$

$\mathbf{x^2 - 12x +39 =2y}$

Divide through by 2

$\mathbf{y = \frac{1}{2}(x^2 - 12x +39)}$

Express 39 as 36 + 3

$\mathbf{y = \frac{1}{2}(x^2 - 12x +36 + 3)}$

Factor out 3/2

$\mathbf{y = \frac{1}{2}(x^2 - 12x +36) + \frac 32}$

Expand the bracket

$\mathbf{y = \frac{1}{2}(x^2 - 6x - 6x +36) + \frac 32}$

Factorize

$\mathbf{y = \frac{1}{2}(x(x - 6) - 6(x -6)) + \frac 32}$

Factor out x - 6

$\mathbf{y = \frac{1}{2}((x - 6) (x -6)) + \frac 32}$

Express as squares

$\mathbf{y = \frac{1}{2}(x - 6)^2 + \frac 32}$

Hence, the equation of the parabola is: $\mathbf{y = \frac{1}{2}(x - 6)^2 + \frac 32}$

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