how do you find the coordinates of the focus of the parabola? (1 Viewer)

Ellie999 · Jul 3, 2007

...im stuck with this question:　find the coordinates of the focus of the parabola 12y = x^2-6x-3
thanx in advance

PC · Jul 3, 2007

12y = x² – 6x – 3

Now complete the square.

12y + 3 = x² – 6x
12y + 3 + 9 = x² – 6x + 9
12y + 12 = (x – 3)²
12(y + 1) = (x – 3)²
4.3.(y + 1) = (x – 3)²

Therfore vertex is (3,-1) and focal distance is 3 units.
The focus is (3,-1+3) = (3,2).

Ellie999 · Jul 4, 2007

thanks PC~

Forbidden. · Jul 4, 2007

PC said:
12y = x² – 6x – 3

Now complete the square.

12y + 3 = x² – 6x
12y + 3 + 9 = x² – 6x + 9
12y + 12 = (x – 3)²
12(y + 1) = (x – 3)²
4.3.(y + 1) = (x – 3)²

Therfore vertex is (3,-1) and focal distance is 3 units.
The focus is (3,-1+3) = (3,2).

Now that's some good revision for the trials.
The methods underline in bold help a LOT, like in Mathematical Induction (Maths. Ext. 1 only)

Annum · Jan 26, 2012

Can anyone help me with y=2x^2+4x-1
I have to find the coordinates of the vertex and focus.
Thankyou sooo much!

SpiralFlex · Jan 26, 2012

Annum said:
Can anyone help me with y=2x^2+4x-1
I have to find the coordinates of the vertex and focus.
Thankyou sooo much!

$You have do it using the axis of symmetry formula or by completing the square. Completing the square is a very powerful technique so I will show you that instead.$

$y=2(x^2+2x-\frac{1}{2})$

$y=2(x^2+2x+1-1-\frac{1}{2})$

$y=2[(x+1)^2-\frac{3}{2}]$

$y=2(x+1)^2-3$

$$We have the parabola in the form of$\; y=a(x-h)^2+k$

$$And the vertex is equal to\;$ (h,k)$

$In this case our vertex is,$

$(-1,-3)$

$Now our second job is to determine it's concavity. Since in the form$ \;ax^2+bx+c, $\;the coefficient of\;$ x^2 \;$is positive, the parabola is concave up.$

$Your pretty diagram should look like this.$

$Now to find the focus.$

$We know the in this case the focus has to be above the vertex. So we would expect the focus to have the coordinates (-1, something).$

$Placing this in the form of\;$ 4a(y-k)=(x-h)^2. $\;Where\;$ a $\;is the focal length.$

$Now,$

$y+3=2(x+1)^2$

$\frac{1}{2}(y+3)=(x+1)^2$

$4a=\frac{1}{2}$

$\therefore a =\frac{1}{8}$

$This is a length. Not a coordinate.$

$So going up 0.125 units from the y coordinate of the vertex will give us it's position.$

$$Hence the position is$ (-1, - \frac{23}{8})$

bleakarcher · Jan 26, 2012

I got the focus to be (-1,-23/8)

nightweaver066 · Jan 26, 2012

bleakarcher said:
I got the focus to be (-1,-23/8)

I got Spiral's answer.

$y = 2x^2 + 4x -1$

$y + 1 = 2(x + 1)^2 - 2$

$y + 3 = 2(x + 1)^2$

$$Vertex: $ (-1, -3)$

$$Focal Length = $ \frac{1}{2}$

$$Focus: $ (-1, -3 + \frac{1}{2})$

$= (-1, -2.5)$

You forgot to account for the 2 in front of the (x + 1)^2

bleakarcher · Jan 26, 2012

I dont know dude, wolframalpha seems to agree wif me too http://www.wolframalpha.com/input/?i=y=2x^2+4x-1

bleakarcher · Jan 26, 2012

click properties

bleakarcher · Jan 26, 2012

so whats going on?

SpiralFlex · Jan 26, 2012

I'm here. What do ya need?

bleakarcher · Jan 26, 2012

wolframalpha says the focus is (-1,-23/8) which is what I too calculated. What I am thinking is you made a mistake where it you said
y-k=4a(x-h)^2 when you meant (x-h)^2=4a(y-k)?

SpiralFlex · Jan 26, 2012

Yeah I think I know what I did wrong. >_> My bad. Lol me and nightweaver swapped it by accident. LOL!

bleakarcher · Jan 26, 2012

yeh i thought so. now if only we knew what on earth nightweaver did LOL.

bleakarcher · Jan 26, 2012

lulz.

SpiralFlex · Jan 26, 2012

I should have used the focus form to check, but was strapped for time.

$(\frac{-b}{2a}, {c+\frac{1-b^2}{4a})$

how do you find the coordinates of the focus of the parabola? (1 Viewer)

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