name the coordinates of the vertex and the focus of the parabola... 3y^2-8x-18y-5=0.. do this by completeing the square. show work please!!!! thanks!!!!!

Question

anonymous · Answer

where did you get 14 on the right side of the equation?

campbell_st · Answer

by completing the square... to get the perfect square I needed to add 9 to both sides on the equation,,,,

but I just realised the mistake...

anonymous · Answer

ya for some reason I thought id have to multlpy 3and 9 and add it to 5 to get 32

campbell_st · Answer

ok... the vertex form I use is 

$$(y - k)^2 = 4a(x - h)$$

vertex is (h, k) and a is the focal length

rewrting you get 

$$3y^2 - 18y = 8x + 5$$

add 27 to both sides of the equation will allow you to complete the square in y

$$3y^2 - 18y + 27 = 8x + 5 + 27$$

which becomes

$$3(y^2 - 6y + 9) = 8x + 32$$

factoring and you get

$$3(y - 3)^2 = 8(x + 4)$$

divide both sides by 3

$$(y - 3) = 4 \times \frac{2}{3}(x + 4)$$

so the vertex is 

(-4, 3)   and the focal length is 2/3 

which means the focus is 

(-4 + 2/3, 3)

hope it helps

anonymous · Answer

@campbell_st  how would I graph this..

campbell_st · Answer

well it looks like

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