SOLUTION: give the standard equation of (3x-2)^2=84y-12. Provide it's vertex, focus, directrix, axis of symmetry, latus rectum and endpoints of latus rectum.

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: give the standard equation of (3x-2)^2=84y-12. Provide it's vertex, focus, directrix, axis of symmetry, latus rectum and endpoints of latus rectum.      Log On


   

Question 1198612: give the standard equation of (3x-2)^2=84y-12. Provide it's vertex, focus, directrix, axis of symmetry, latus rectum and endpoints of latus rectum.
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
we use the standard form
%28x-h%29%5E2=4p%28y-k%29+
for parabolas that have an axis of symmetry parallel to the y-axis
%283x-2%29%5E2=84y-12
9x%5E2+-+12x+%2B+4=84y-12
9x%5E2+-+12x+=84y-12-4
9%28x%5E2+-+%2812%2F9%29x%29+=84y-16
9%28x%5E2+-+%284%2F3%29x%2B%284%2F6%29%5E2%29+-9%284%2F6%29%5E2=84y-16
9%28x%5E2+-+%284%2F3%29x%2B%282%2F3%29%5E2%29+-9%282%2F3%29%5E2=84y-16
9%28x+-+2%2F3%29%5E2+-4=84y-16
9%28x+-+2%2F3%29%5E2+=84y-16%2B4
9%28x+-+2%2F3%29%5E2+=84y-12
9%28x+-+2%2F3%29%5E2+=84%28y-12%2F84%29
9%28x+-+2%2F3%29%5E2+=84%28y-1%2F7%29
%28x+-+2%2F3%29%5E2+=%2884%2F9%29%28y-1%2F7%29
%28x+-+2%2F3%29%5E2+=%2828%2F3%29%28y-1%2F7%29

compare to %28x-h%29%5E2=4p%28y-k%29+

4p=28%2F3
p=28%2F12
p=7%2F3
h=2%2F3
k=1%2F7

so
vertex is at (h,+k)=(2%2F3,+1%2F7)
focus:(h,+k%2Bp)=(2%2F3, 1%2F7%2B7%2F3)=(2%2F3, 52%2F21)
directrix: y+=k-p=1%2F7-7%2F3=+-46%2F21
axis of symmetry: x=h so x=2%2F3
latus rectum:4p=+28%2F3
endpoints of latus rectum: (h%2B2p,+k%2Bp) and (h-2p,+k%2Bp)
(h%2B2p,k%2Bp) =(2%2F3%2B2%287%2F3%29, 1%2F7%2B7%2F3)=(16%2F3, 52%2F21)
and
(h-2p,+k%2Bp)=(2%2F3-2%287%2F3%29,+1%2F7%2B7%2F3)=(-4, 52%2F21)