Question

What is the equation in standard form of the parabola with a focus at (14,5) and a directrix of y= -15?

Answer 1

The equation of parabola is `y=1/40(x-14)^2-5`

Explanation:

Focus is at `(14,5)`and directrix is `y=-15`. Vertex is at midway

between focus and directrix. Therefore vertex is at

`(14,(5-15)/2) or (14, -5)` . The vertex form of equation of

parabola is `y=a(x-h)^2+k ; (h.k) ;` being vertex. Here

`h=14 and k = -5` So the equation of parabola is

`y=a(x-14)^2-5`. Distance of vertex from directrix is

`d= 15-5=10`, we know `d = 1/(4|a|):. |a|=1/(4d)` or

`|a|=1/(4*10)=1/40` . Here the directrix is below

the vertex , so parabola opens upward and `a` is positive.

`:. a=1/40` Hence the equation of parabola is

`y=1/40(x-14)^2-5`
graph{1/40(x-14)^2-5 [-90, 90, -45, 45]} [Ans]

Answer 2

`(x-14)^2=40(y+5)`

Explanation:

`"the standard form of a parabola in "color(blue)"translated form"` is.

`•color(white)(x)(x-h)^2=4p(y-k)`

`"where "(h,k)" are the coordinates of the vertex"`

`"and p is the distance from the vertex to the focus"`

`"since the directrix is below the focus then the curve"`
`"opens upwards"`

`"coordinates of vertex "=(14,(5-15)/2)=(14,-5)`

`"and "p=5-(-5)=10`

`rArrrArr(x-14)^2=40(y+5)larrcolor(red)"equation of parabola"`