Question

What are the coordinates of the focus of the conic section shown below?

y^2-4x+4y-4=0

Answer 1

First of all you would want to express this equation in its Standard Form:\[f(x) =a(x \pm h)^{2}\pm k\] for which \[(\pm h, \pm k)\] expresses the vertex of the parabola and \[a\] the amplitude. But to do this, you must COMPLETE THE SQUARE, using the little formula \[C = (b/2)^{2}\] where \[f(x) =ax ^{2}+bx+c\] and \[C\] is the value to complete the square. Now lets start using the example above:\[y ^{2}-4x+4y-4=0\] can be rearranged to \[y ^{2}+4y-4=-4x\] which in turn is\[(-1/4)y ^{2}-y+1=x\]. Now that we have it in the General Form, lets complete the square to have the Standard Form. Using the little formula. First you have to factor the amplitude from both of the variables, in this case\[a=(-1/4)\]. Therefore we have that\[x = (-1/4)(y ^{2}+4y)+1\] Now lets use the little formula:\[C=(b/2)^{2}=(4/2)^{2}=(2)^{2}=4\]. This number is the key to completing the square but since you can't put numbers in equations without a toll, you can only put it if doesn't change the original equation. Therefore it must add zero. \[x = (-1/4)(y ^{2}+4y +4 -4)+1\]. Now you always want the positive C, in this case +4. But -4 you must first multiply it with the amplitude:\[x = (-1/4)(y ^{2}+4y+4)+(-1/4)(-4)+1 = (-1/4)(y ^{2}+4y+4)+1+1=\]\[(-1/4)(y ^{2}+4y+4)+2\]. Now the square would be complete with the number you have before squaring in the little formula\[(b/2)=(4/2)=2\]. In this case you will have:\[x = (-1/4)(y ^{2}+4y+4)+2=(-1/4)(y+(4/2))^{2}+2=(-1/4)(y+2)^{2}+2\]. And there you have it, the Standard Form:\[f(y) =a(y \pm h)^{2} \pm k=(-1/4)(y+2)^{2}+2\]. Now as I mention earlier, the vertex is (\[(\pm h, \pm k)\] but you must always change the sign for \[h\] when writing the vertex, therefore you have that the vertex for our example is\[v =(-2,2)\]. FINALLY WE DID IT...Not exactly. Now you must find the coordinate for the focus. To do this, we need another formula:\[ p = 1/4a\]. Now remember that a is the amplitude. Therefore we have that \[p= 1/4a=1/4(-1/4)=1/(-1)=-1\]. Now this p is the distant you have from the vertex to the focus. Therefore you must always SUM p to the vertex to find the focus. Always remember that the focus is inside the parabola, therefore in the case of our example we have a parabola that opens to the right, so you must sum P to the x -axis of the vertex since the focus will be inside the parabola. Therefore \[f=(-h+p,k)=(-2+(-1),2)=(-3,2)\]. And there you have it. Peace, Love and Happiness from Puerto Rico.

Answer 2

Its actually (-1,-2) ..