Question

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

Answer 1

Answer-

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

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,

$=\sqrt{(3-3)^2-(9-5)^2}\\\\=\sqrt{4^2}\\\\=4$

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

then the equation becomes,

$(x-3)^2=4\times 4(y-5)\\\\(x-3)^2=16(y-5)$

Answer 2

Answer:

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

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