Graph the parabola 8 ( x + 1 ) = ( y − 2 ) 2 8\left(x+1\right)=\left(y-2\right)^2 8 ( x + 1 ) = ( y − 2 ) 2 , and find the focus point and directrix line.

Welcome back everyone. Let's give this problem a try. So in this problem, we are asked to graph the parabola eight times x plus one equals y minus two squared, and find the focus point and directrix line. Now what I notice with this problem is we're dealing with a situation where the y variable is squared. And because of this, that means we're going to have some kind of sideways parabola on our graph. So what I can do is use the standard equation for a horizontal parabola, which is four p times x minus h is equal to y minus k squared. Now from here I can extract the vertex of our parabola, and I noticed that this h corresponds with one. But since we have a plus sign rather than a minus sign, that means h is going to be negative one. Now I can see that this k corresponds with two, and we have a minus sign there and there so this checks out. So all we need is the positive two for k. So the vertex of our parabola is going to be at negative one two, which on our graph we go back to negative one and up two will be right there. So that's the vertex. Now what I also need to do is find the focus and directrix and then graph the parabola. So to find the directrix and focus, what I can do is use this four p. I can set four p equal to everything in front of the x minus h argument, which is gonna be this eight. So we're going to have that four p is equal to eight. I can divide four on both sides, and eight divided by four is two. So that means our p value is equal to two. Now Now notice we got a positive value. And because we got a positive p value, that means our parabola is going to open to the right. Now this also tells us where our focus and directrix are. The focus points starting at the vertex is going to be two units to the right, because that's the direction that the parabola opens. So we can see that our focus point is going to be at the point one comma two, and that's our focus. Now what we need to find next is the directrix line, and the directrix line is opposite the direction that the parabola opens at a distance of p. So that means we're going to go back one, two units to get our directrix line. So our directrix line is going to be at x equals negative three, and this would be our directrix. And then what we found for the point over here would be our focus. So we have found the focus and directrix in this problem, and our last step is going to be to actually graph this parabola. And to graph the parabola, we can use the vertex, and we can go two p units up and down to find the width. So two p is the same thing as two times two, which is four. So I can go up one, two, three, four units, which is actually gonna be somewhere up here. So we're actually past what we have on our graph here. It's gonna be if we extend this up, it would be at six. So we go four units up, and then we can go one, two, three, four units down. And what this is going to do is give us a parabola, which looks something like that. So this would be our parabola, and this would be the focus and directrix associated with this parabola as well. So that is how you can solve this problem. Hope you found this helpful. Let me know if you have any questions.

16. Parametric Equations & Polar Coordinates

Conic Sections

Calculus

16. Parametric Equations & Polar Coordinates

Conic Sections

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Multiple Choice

Graph the parabola $8\left(x+1\right)=\left(y-2\right)^2$ , and find the focus point and directrix line.

option a

option b

optiuon c

option d

Verified step by step guidance

Step 1: Recognize the given equation of the parabola, 8(x+1) = (y-2)^2, and identify its standard form. This equation represents a parabola that opens horizontally because the squared term is on the y-variable.

Step 2: Rewrite the equation in the standard form for a horizontally opening parabola: (y-k)^2 = 4p(x-h), where (h, k) is the vertex and p is the distance from the vertex to the focus or directrix. In this case, rewrite the equation as (y-2)^2 = 8(x+1).

Step 3: Identify the vertex of the parabola. From the equation (y-2)^2 = 8(x+1), the vertex is at (-1, 2). This is derived from the values of h and k in the standard form.

Step 4: Determine the value of p, which is the distance from the vertex to the focus or directrix. From the equation, 4p = 8, so p = 2. Since the parabola opens to the right, the focus is located at (h+p, k), which is (-1+2, 2) = (1, 2). The directrix is a vertical line given by x = h-p, which is x = -1-2 = -3.

Step 5: Graph the parabola using the vertex (-1, 2), the focus (1, 2), and the directrix x = -3. Draw the curve opening to the right, ensuring it is symmetric about the line passing through the vertex and focus.