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Vadim26 [7]
3 years ago
14

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

Mathematics
2 answers:
Kay [80]3 years ago
8 0

<u>Answer-</u>

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

<u>Solution-</u>

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)

Kruka [31]3 years ago
7 0

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).



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