Question

Refer to your answers to the questions from Part 1 of Project 1.

Any point on the parabola can be labeled (x,y), as shown.
What are the distances from the point (x,y) to the focus of the parabola and the directrix?
SELECT TWO ANSWERS.

Answer 1

`\textsf{Distance\;to\;the\;focus:\;$√((x-2)^2+(y+4)^2)$}`

`\textsfy+6`

Explanation:

Distance to the focus

The focus of a parabola is a fixed point located on the axis of symmetry, such that all points on the parabola are equidistant to both the focus and the directrix. It is located inside the parabola. Therefore, the focus of the given parabola is (2, -4).

To find the distance from a point (x, y) to the focus of the parabola, we can use the distance formula:

`\boxed{\begin{minipage}{7.4 cm}\underline{Distance Formula}\\\\$d=√((x_2-x_1)^2+(y_2-y_1)^2)$\\\\\\where:\\ \phantom{ww}$\bullet$ $d$ is the distance between two points. \\\phantom{ww}$\bullet$ $(x_1,y_1)$ and $(x_2,y_2)$ are the two points.\\\end{minipage}}`

Let the focus be the first point: (x₁, y₁) = (2, -4)

Let (x, y) be the second point: (x₂, y₂) = (x, y)

Substitute the points into the distance formula:

`d=√((x-2)^2+(y-(-4))^2)`

`d=√((x-2)^2+(y+4)^2)`

Therefore, the distance to the focus is:

`\large\boxed{√((x-2)^2+(y+4)^2)}`

`\hrulefill`

Distance to the directrix

The directrix of a parabola is a fixed line outside of the parabola that is perpendicular to the axis of symmetry. Therefore, the directrix of the given parabola is y = -6.

The distance from any point on the parabola to the directrix is the absolute value of the difference between the y-coordinate of the point and the y-coordinate of the directrix. Therefore:

`d = |y - (-6)|`

`d = |y + 6|`

Therefore, the distance to the directrix is:

`\large\boxed`

Answer 2

distance to thefocus=
`\tt √(((x-2)^2 + (y+4)^2)`

Distance to the directrix = |y + 6|

Explanation:

Given:

vertex(h,k)=(2,-5)

focus(h,k+p)=(2,-4)

directrix(y)=

Solution:

The distance between the point (x,y) and the focus(2,-4) is:

distance =
`\tt √((x_2-x_1)^2+(y_2-y_1)^2)`

distance =
`\tt √((x-2)^2 + (y-(-4))^2)`

distance to the focus=
`\tt √(((x-2)^2 + (y+4)^2)`

We have:

the distance from the point (x, y) to the directrix is given by:

Distance to directrix = |y - (k - 1)|

Distance to directrix = |y - (-5 - 1)|

Distance to directrix = |y + 6|

Therefore, the distance between the point (x,y) and the directrix is:

Distance to the directrix = |y + 6|