Need help writing equation of parabola?

Vertex (7, 5):

h = 7

k = 5

Directrix:

x = 1

Since the directrix is to the left of the vertex, the parabola will be horizontally-oriented and opening to the right.

p = h - x:

p = 7 - 1

p = 6

a = 1 / 4p:

a = 1 / 4(6)

a = 1/24

Equation:

x = a(y - k)² + h:

x = 1/24 (y - 5)² + 7

x = 1/24 (y² - 10y + 25) + 7

x = 1/24 y² - 5/12 y + 25/24 + 7

x = 1/24 y² - 5/12 y + 25/24 + 168/24

x = 1/24 y² - 5/12 y + 193/24

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or

x = 0.0417y² - 0.4167y + 8.0417

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Standard equation:

y= a(x-h)^2 + k {(h,k) is the vertex}

Re-written as:

(x – 7)^2 = 4p(y – 5) {4p = 1/a}

Assuming Directrix Y=1,

Distance between vertex and directrix = absolute value (1-5) = +4 = p

(x-7)^2 = 4(4)(y-5)

y-5 = [(x-6)^2]/16

y= [(x-6)^2]/16 + 5

Directrix cannot = 1

It can be x = 1 or y = 1

Y = x^2 - 6x - 1 ==> y = (x^2 - 6x + 9) - 1 - 9 ==> y = (x - 3)^2 - 10 EDIT: relying on the learn, the typical types of parabolas do differ. In this case, the reply is A. Hope this helps!

(x-7)^2 +(y-5)^2 =(x-1)^2

x^2-14x +49 +y^2-10y +25 =x^2-2x+1

y^2-10y +25 =12x-48

(y-5)^2 =12x-48

12x =(y-5)^2+48

x = (1/12)(y-5)^2 +4 ...............Ans