Need help with conics!!

<p>So we did a week of conics and I was sick all that week. Now we have a review for the final and I don’t get it, plus it’s not in our book! Lol help please??</p>

<li><p>Write the y = 3x^2 - 6x + 2 in standard form (I thought this was standard form?)</p></li>
<li><p>Identify the coordinates of the vertex and focus, the equations of the axis of symmetry and directrix, and the direction of opening of the parabola with the equation y^2 - 8y + 18 = x. </p></li>
<li><p>Graph this 9x^2 + 4y^2 = 36. (Just need to know what kind of shape this is. Eclipse?)</p></li>
</ol>

<p>Thanks so much!</p>

<ol>
<li>Standard form for quadratics is f(x) = a(x-h)^2 + k. Complete the square and you'll be able to put it in this form.</li>
<li>Too lazy. Sorry. :D</li>
<li>Divide both sides by 36 to get x^2/4 + y^2/9 = 1. This is a vertically stretched ellipse.</li>
</ol>

<p>Thanks for 1 & 3. Anybody know 2? [I hate this conic crap.]</p>

<p>subtract 18 to the other side y^2-8y= x-18
Complete the square for y (y-4+16)=x-18+16
(y-4)^2= x-2
The normal equation for parabolas is (x-h)^2= 4a(y-k) your problem however switches the x's and y's so instead of facing up and down it will face left and right. and it's positive( I believe) so it will face right. the vertex is (h,K) so it (2,4) to find the focus you need to solve for a and since its usually 4a a will be 4 since you don't have a number outside the x, this also gives you the distance for the directrix from the outside of the vertex. I hope all of this is correct and I hope it helps I'm not a good explaner.</p>

<p>Thanks. I knew it was sideways but I couldn't remember how to get there.</p>