Question

Write the equation of the parabola with a vertex (3,5) and a focus of (3,9)

Answer 1

Given: Vertex (3,5) and focus (3,9) of the parabola.

As you can see that the x-coordinates of the vertex and focus are the same, so, this is a regular vertical parabola.

The Standard form of the equation of the parabola is;
`(x-h)^2 = 4a(y-k)` , where a≠0. ......[1]

The vertex of this parabola is at (h, k).

and the focus of this parabola is at (h, k+a).

From the given,

h = 3, k=5 and k+a = 9

To solve for a;

k+a = 9

Substitute the value of k =5 in above expression:

5+a =9

⇒ a= 9-5 =4

Therefore, the vertex and focus are 4 units apart.

Hence, the equation of parabola by substituting the value of h, k and a in equation [1] ;

`(x-3)^2=4\cdot 4(y-5)` or

`(x-3)^2=16(y-5)`.

Answer 2

`\boxed{\boxed{(x-3)^2=16(y-5)}}`

Solution-

Given,

the parabola has vertex at (3,5) and focus at (3,9)

It can be easily noticed that focus and vertex lie on the same vertical line,
`x=3`

Hence, the axis of symmetry is a vertical line ( a line perpendicular to x-axis). Also, the focus lies to the top of the vertex so the parabola will open up upwards.

General form of parabola vertex at (h, k)and vertical line of symmetry is,

`(x-h)^2=4a(y-k)`

Putting the values,

`(x-3)^2=4a(y-5)`

We can find the value of 'a' which is distance between vertex and focus by using distance formula,

`=√((3-3)^2-(9-5)^2)\\\\=√(4^2)\\\\=4`

Taking only +ve value, as the parabola opens upwards.

then the equation becomes,

`(x-3)^2=4* 4(y-5)\\\\(x-3)^2=16(y-5)`