In Exercises 5–16, find the focus and directrix of the parabola w...

Question

In Exercises 5–16, find the focus and directrix of the parabola with the given equation. Then graph the parabola.y^2 = 16x

Accepted Answer

Hey, everyone for the following equation of a Parabola we are asked to solve for the focus and the direct tricks. And then graph, the given equation is Y squared is equal to X. Here we have four answer choice options. Each of which display a graph including the parabola and the direct tricks. And each solution states the focus and the direct tricks. So beginning to solve this problem again, we are given the equation of a Parabola as Y squared is equal to 24 X. And we see that this equation matches the form where we have the quantity of Y minus K squared is equal to four A times the quantity of X minus H. So here the first step is to solve for or identify the value for A. And so we just need to compare this formula with our given equation. So we see that the coefficient of X in this case is 24. So we can equate 24 with four A from our formula. So we have 24 is equal to four A. And now both sides by four, we see that the value for A is just going to be six. And so now we can move on to identifying our vertex. So if we recall the vertex is at the point H K, and now looking at our given equation, we see that the vertex is going to be at the origin or at zero and zero. And so now we can also identify our focus. Well, the focus is going to be at the point H plus A for the X coordinate and then at K for the Y coordinate. And so now just substituting in our information that we have, we have zero plus C for the X coordinate and just zero for the Y coordinate. So we see after simplifying that the focus is going to be at the 00.6 and zero. And now we also can note that the direct tricks and the focus lie at equal distances from the vertex of the parabola. And the equation of the direct can be calculated based on the equation of the Parabola. So since the vertex is at the origin or at 00, and the focus is at a, then the equation of the direct is going to be at X is equal to negative A. Well, at negative A, we have the negative quantity of negative the negative quantity of positive six. And so now simplifying we see that our direct tricks is going to be for when X is equal to negative six. So now with our direct tricks identified as well as our vertex and the focus we can now graph each of these components. And so we see that we have a graph here now of our Parabola which opens facing the right side of our graph. And now we have a vertical line, a vertical dash line representing our directs at X is equal to negative six. We have our vertex at the origin and our focus at six and zero. So with this graph and now comparing our solutions here with our answer choice options, we see we are left with answer choice. A where again our focus is at six and zero and our direct tricks is at X is equal to negative six. Thanks for watching. I hope you found this video helpful and I'll see you again next time.

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In Exercises 5–16, find the focus and directrix of the parabola with the given equation. Then graph the parabola.y^2 = 16x

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