Question

Some Algebra help, please?

I need to find the axis of symmetry, vertex, p, focus, and directrix to these problems.

1. 3y+24y-3x+42=0
2. x^2+2x+12y+3y=0
3. y^2-4x+2y+5=0
4. y^2-2x-2y+5=0

I've tried #4 by myself but couldn't get pass: y^2-2y=2x-5.

Any help is appreciated!

Answer 1

The first equation doesn't feature the square of either x or y, so I'll simply ignore it.

The second equation does have the term x^2, leading me to conclude that this represents a vertical parabola. (Why?)

My next move would be to use Completing the Square to come up with an expression in the form (x-h)^2; that h would be the x-coordinate of the vertex. I could not Complete the Square for the y-terms, because there's no y^2.

Our first goal is to rewrite the second equation in the form y-k = a(x-h)^2, because this reveals the coordinates of the vertex, (h,k).

In the end, we need to transform this equation for the parabola into one of the other standard forms: 4py=x^2, or, in this case, because the vertex isn't at (0,0),

4p(y-k)=(x-h)^2.

Review what p represents: it's the distance from the vertex to the focus. if you know the coordinates of the vertex, and know p, you can then find the coordinates of the focus.

Good luck!

Answer 2

I'll guess that the first problem is supposed to be \[3y^2+24y-3x+42=0\]If it isn't, well, too bad, this is the problem I'm doing!

We can factor out a 3 from every term here, and that will make life easier, so let's do so.
\[3(y^2+8y-x+14)=0\]Now we can divide both sides by 3\[y^2+8y-x+14=0\]We have a term in \(y^2\) on the left, but no corresponding term in \(x^2\), so let's move the \(x\) term to the right hand side by adding \(x\) to each side:

\[y^2+8y-x+x + 14 = 0+x\]\[y^2+8y+14 = x\]Now, to complete the square, we want to bundle up all the terms containing \(y\) in a neat little package which will look like \((y+4)^2\). To do that, however, we need to find out what the proper third term in the expand of \((y+4)^2\) would be.

If we have a perfect square, \((y+a)^2\) we can expand it like this:
\[(y+a)(y+a) = y(y+a) + a(y+a) = y^2 + ay + ay +a^2\]\[\qquad=y^2+2ay + a^2\]What this means to us is that if we line up our \(y\) and \(y^2\) terms next to that, we can figure out what the constant term we need is.

\[y^2 + 8y\]\[y^2+2ay\]For those to be equal, \[8y=2ay\]\[8=2a\]\[a=4\]and \[(y+4)^2 = y^2 + 2*4*y + 4*4 = y^2 + 8y + 16\]

The result is that we can take \(y^2 + 8y + 16\) and turn it into our perfect square. But what if we don't have a spare \(16\) hanging around to use? Not a problem! We'll just add one, and take it away at the same time so that we don't disturb our equation.

Watch my hands carefully :-)

\[y^2+8y + 14 = x\]\[(y^2 + 8y ) + 14 = x\]\[(y^2 + 8y + 16 - 16) + 14 = x\]Now \(16-16=0\) so the value is still the same, but it's in a more convenient form.\[(y^2+8y+16) - 16 + 14 = x\]\[(y+4)^2 -16 + 14 = x\]\[(y+4)^2 -2 = x\]\[(y+4)^2 = x-2\]

So to complete the square, we took half the value of the coefficient of \(y\), squared it, and simultaneously added and subtracted it. The part we added got sucked into our completed square, and the part we subtracted preserved the balance of our equation. Depending on what you are doing, it may be more convenient to instead add the same value to both sides of the equation; the two operations are equivalent, so choose whatever is most convenient and comfortable for you.