determine  the focus, vertex and directrix of the parabola of x^2+6x-8y+17=0
in Geometry Answers by

Your answer

Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
To avoid this verification in future, please log in or register.

1 Answer

x²+6x-8y+17=0,

x²+6x+9-9-8y+17=0,

(x+3)²-9-8y+17=0,

(x+3)²-8y+8=0,

(x+3)²/8-y+1=0,

y-1=(x+3)²/8, so vertex is at (-3,1).

The coefficient of the squared term is 1/8=1/4f where f is the vertical distance of the focus and directrix line from the vertex, so 4f=8, f=2. The focus is “inside” the parabola, so y=1 at the vertex, and the focus is at y=1+2=3 on the axis of symmetry x=-3, so its coords are (-3,3).

The directrix is “outside” the parabola at y=1-2=-1, the line y=-1.

by Top Rated User (764k points)

Related questions

1 answer
asked Oct 8, 2014 in Algebra 2 Answers by baruchqa Level 1 User (280 points) | 200 views
1 answer
asked Oct 8, 2014 in Algebra 2 Answers by baruchqa Level 1 User (280 points) | 162 views
Welcome to MathHomeworkAnswers.org, where students, teachers and math enthusiasts can ask and answer any math question. Get help and answers to any math problem including algebra, trigonometry, geometry, calculus, trigonometry, fractions, solving expression, simplifying expressions and more. Get answers to math questions. Help is always 100% free!
85,099 questions
90,236 answers
2,058 comments
60,541 users