How do you find the vertex, focus and directrix of x^2 + 6x + 8y + 25 = 0 x2+6x+8y+25=0?

1 Answer
Douglas K.
Jan 30, 2017

When given the equation of a parabola, y(x)=ax^2+bx+cy(x)=ax2+bx+c:
h=-b/(2a)h=−b2a
k=y(h)k=y(h)
f=1/(4a)f=14a
The vertex is: (h,k)(h,k)
The focus is: (h,k+f)(h,k+f)
The equation of the directrix: y =k-fy=k−f

Explanation:

Write the given equation in the form, y(x) = ax^2+bx+cy(x)=ax2+bx+c:

x^2+6x+25+8y=0x2+6x+25+8y=0

x^2+6x+25=-8yx2+6x+25=−8y

y(x) = -1/8x^2 -3/4x-25/8y(x)=−18x2−34x−258

a = -1/8, b = -3/4, and c = -25/8a=−18,b=−34,andc=−258

Compute the value of h:

h = -b/(2a)h=−b2a

h = -(-3/4)/(2(-1/8))h=−−342(−18)

h = -3h=−3

Compute the value of k:

k = y(h)k=y(h)

k = y(-3)k=y(−3)

k = -1/8(-3)^2 - 3/4(-3)-25/8k=−18(−3)2−34(−3)−258

k = -2k=−2

Compute the value of f:

f = 1/(4a)f=14a

f = 1/(4(-1/8))f=14(−18)

f = -2f=−2

The vertex is: (-3,-2)(−3,−2)
The focus is: (-3,-4)(−3,−4)
The equation of the directrix is y = 0y=0

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