Question

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

f(x) = −(x + 2)2 − three fourths

f(x) = (x + 2)2 + three fourths

f(x) = −(x − 2)2 + three fourths

f(x) = −(x − 2)2 − three fourths

Answer 1

`y = - (x - 2)^2 - (3)/(4)`

Explanation:

The parabola must have all the points to be equally distant from the point of focus and the directrix, where directrix a horizontal line and the focus being a point given.

To derive an equation for the parabola, we will use the formula of distance. Now we have a random point on the parabola (x,y) and the point (x,y) will be equidistant from the focus and the directrix. So using the distance formula, we will get:

`\sqrt{(y -((-1)/(2)}))^2} = √((x-2)^2 + (y - (-1)^2)`

The square root of y-(-1/2) from the directrix, and the righthand side of the equal sign is derived from the focus point. Now simplify it to get:

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

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

`-y - (3)/(4) = x^2 - 4x + 4`

`-y = (x - 2)^2 + (3)/(4)`

`y = - (x - 2)^2 - (3)/(4)`

.

Answer 2

Focus of the parabola is (2,-1) and directrix is y = -1/2.

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

According to definition of parabola, the distance between point (x,y) and focus (2,-1) would be same as the distance between the point (x,y) and directrix y = -1/2.

`√((x-2)^2+(y+1)^2) = \sqrt{(y+(1)/(2)) ^2} \\\\(x-2)^2+(y+1)^2=(y+(1)/(2) )^2 \\\\(x-2)^2 + y^2 + 2y +1 = y^2 +y +(1)/(4) \\\\(x-2)^2 =  -y^2 - 2y -1 + y^2 +y +(1)/(4) \\\\(x-2)^2 =  -y -1 +(1)/(4) \\\\(x-2)^2 =  -y +(-4+1)/(4) \\\\(x-2)^2 =  -y +(-3)/(4) \\\\y =  -(x-2)^2  +(-3)/(4) \\\\`

Hence, option D is correct, i.e. f(x) = −(x − 2)² − three fourths.