Sorry asking for so much help

**jhonwashington** · November 24th 2006, 11:39 AM

I have more questions, sorry to bother

they ask me to write the equation in standard form, then find the center and radius x^2+y^2+8x+7= 0

they also ask me to find the vertex, focus and the lenght of the lactus rectum 8y=x^2

and just one last, so sorry to ask for that much help

find the equation of the parabola that has a focus at (-3,4) and directriz y=2

thank you.

**earboth** · November 24th 2006, 09:37 PM

Originally Posted by jhonwashington

I have more questions, sorry to bother

they ask me to write the equation in standard form, then find the center and radius x^2+y^2+8x+7= 0

...

Hello,

you have: $x^2+y^2+8x+7= 0$ . Now accomplish the squares:
$x^2+8x+16+y^2=-7+16$
$(x+4)^2+y^2=9$ . This equation describes a circle with the centre M(-4, 0) and the radius r = 3.

EB

**earboth** · November 24th 2006, 09:57 PM

Originally Posted by jhonwashington

...

they also ask me to find the vertex, focus and the lenght of the lactus rectum 8y=x^2...

Hello,

the general form of your parabola is: $x^2 = 2p \cdot y$ . This is a parabola which opens up and has it's vertex at V(0, 0). The focus has the (general) coordinates $F\left(0,\frac{p}{2}\right)$ . With your equation it becomes: $F\left(0,2\right)$ .
The lactus rectum is the chord passing through the focus and perpendicular to the (main) axis. With a parabola the lactus rectum has always the length 2p. In your case the lactus rectum is 8.

EB

**earboth** · November 25th 2006, 01:55 AM

Originally Posted by jhonwashington

I have more questions, ...
find the equation of the parabola that has a focus at (-3,4) and directriz y=2.

Hello,

the directriz is below the focus thus the parabola opens upward and the main axis is x = -3.

The midpoint on the main axis between the focus and the directriz is the vertex V(d, c) thus your vertex is V(-3, 3).

The general form of your parabola is:
$(x-d)^2=2p\cdot (y-c)$

The distance between focus and directriz is p.

Plug in the values you know:

$(x+3)^2=2\cdot 2\cdot (y-3)$ . Solve for y:

$y=\frac{1}{4}x^2+\frac{3}{2}x+\frac{21}{4}$

EB

**jhonwashington** · November 25th 2006, 09:30 AM

thanks earboth, I wish I were as smart as you are.

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Sorry asking for so much help

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