Question

How do you find the standard form of the equation of the parabola with the given characteristics vertex (3,1) focus (8,1)? thank you

Answer 1

This is the standard form of a parabola that opens to the right:

`x = 1/20y^2-1/10y+61/20`

Explanation:

Please observe that the focus (at point `(8,1)`) is shifted 5 units to the right of the vertex (at point `(3,1)`); this means that the directrix is a vertical line 5 units to the left of the vertex, `x = -2`

A parabola is defined as the locus of points equidistant from its focus and its directrix.

The distance from the focus, `(8,1)`, to any point `(x,y)` on the parabola is:

`d = sqrt((x-8)^2+(y-1)^2)" [1]"`

The distance from the directrix, x = -2, to any point `(x,y)` on the parabola is:

`d = sqrt((x--2)^2+(y-y)^2)" [2]"`

Because the points must be equidistant, we can set the right side of equation [1] equal to the right side of equation [2]:

`sqrt((x-8)^2+(y-1)^2) = sqrt((x--2)^2+(y-y)^2)`

Square both sides:

`(x-8)^2+(y-1)^2 = (x--2)^2+(y-y)^2`

`(y - y)^2` is always 0 and `--` is +:

`(x-8)^2+(y-1)^2 = (x+2)^2`

Expand the squares:

`x^2-16x+64+y^2-2y+1=x^2+4x+4`

Combine like terms:

`20x = y^2-2y+61`

Divide both sides by 20:

`x = 1/20y^2-1/10y+61/20`