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This is a priority! The equation of a parabola is 1/16 (y+3)^=x+4 what are the coordinates of the focus?

  • carlosego: To solve this problem you must apply the proccedure shown below:

     1. You have the following equation of a parabola:

     1/16(y+3)^2=x+4 

     2. Rewriting the equation of the parabola, you have:

     
    (y+3)^2=16(x+4)

     3. The vertex of the parabola is:

     (h,k)=(-4,-3)

     3. Now, you must find the coefficient of the unsquared part, as below:

     4p=16
     p=16/4
     p=4

     4. As you can see, the y part is squared and p is positive. This means that the parabola that opens to the right.

     5. The focus of the parabola has to be 4 units to the right of the vertex. Therefore, the focus is:

     (0,-3)

     The answer is: (0,-3)
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