Question

What is the vertex form of the equation of the parabola with a focus at (29,37) and a directrix of `y=34`?

Answer 1

The vertex form of equation is `y=1/6(x-29)^2+35.5`

Explanation:

Focus is at `(29,37)`and directrix is `y=34`. Vertex is at midway

between focus and directrix. Therefore vertex is at

`(29,(37+34)/2) or ( 29,35.5)` The vertex form of equation of

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

`h=29 and k = 35.5`, so the equation of parabola is

`y=a(x-29)^2+35.5`. Distance of vertex from directrix is

`d= 35.5-34=1.5`, we know `d = 1/(4|a|)`

`:. 1.5 = 1/(4|a|) or |a|= 1/(1.5*4)=1/6`. Here the directrix is

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

`:. a=1/6` . The equation of parabola is `y=1/6(x-29)^2+35.5`

graph{1/6(x-29)^2+35.5 [-160, 160, -80, 80]} [Ans]

Answer 2

`(x-29)^2=6(y-71/2)`

Explanation:

The vertex of parabola having focus at `(29, 37)` & directrix `y=34` will lie on the axis of symmetry in the middle of focus & directrix whose coordinates ate given as

`Vertex\equiv(\frac{29+29}{2}, \frac{37+34}{2})`

`\equiv(29, 71/2)`

Now, the equation of vertical parabola with vertex at `(x_1, y_1)\equiv(29, 71/2)`will be

`(x-x_1)^2=4a(y-y_1)`

`(x-29)^2=4a(y-71/2)\ ...........(1)`

The equation of directrix of above parabola:

`y-71/2=-a`

`y=71/2-a`

But equation of directrx is given `y=34` hence

`71/2-a=34`

`a=71/2-34`

`a=3/2`

setting value of `a` in (1), we get equation of parabola:

`(x-29)^2=4\cdot 3/2 (y-71/2)`

`(x-29)^2=6(y-71/2)`