write the standard equation of the parabola with focus (2,1) and vertex (-5, 1)

Question

Algebra- Conics Section

Answer 1

Remember the general form for a parabola which opens to the right: 

X-h = 1/4p*(Y-k)²  where the vertex is at (h,k) and the distance from the vertex to the focus is p. 

Thus, h = -5, k = 1, and p = 2-(-5) = 7. Plugging in:

X+5 = 1/(4*7) * (Y-1)² 

This can be simplified a bit. 

X+5 = (Y-1)²/28 

X   = (Y-1)²/28 -5         Now we FOIL out the (Y-1)²

(Y-1)(Y-1) = Y*Y + Y*-1 + -1*Y + -1*-1 = Y² - Y - Y +1 = Y² -2Y +1

       =  (Y²-2Y+1)/28 +1   And we distribute out the 28 to everything in the parentheses

X    = Y²/28 -2Y/28 + 1/28 - 140/28  

X    = Y²/28 -Y/14 + -139/28  

This is the final solution for the equation. Note that what we had earlier,  X = (Y-1)²/28 -5, is also acceptable and may be more helpful in some cases. Both are the same equation, it only depends on what form you want for the answer. 

Edited because I had X and Y mixed up. Remember that if the vertex and the focus have the same Y value, the parabola opens sideways. Sorry for the confusion.

 

Answer 2

The standard equation of the parabola should have the form of y = ax^2 + bx + c, or x = ay^2 + by + c.

Since you are given focus (2,1) and vertex (-5, 1), and you know this parabola opens to the right, you can start with vertex form.

x = (1/4p)(y-1)^2 - 5, where p = 2+5 = 7, the distance from the focus to the vertex.

Simply into standard form,

x = (1/28)(y^2-2y+1) - 5 = (1/28)y^2 - (1/14)y - 4 27/28