In Exercises 17–30, find the standard form of the equation of eac...

Question

In Exercises 17–30, find the standard form of the equation of each parabola satisfying the given conditions.Focus: (0, - 25); Directrix: y = 25

Accepted Answer

Hello. Everyone in this video. We're going to be looking at this practice problem where we want to write the equation of the problem in standard form. Given the conditions that has a focus at zero negative and direct tricks at Y. Is equal to 10. So in order to get started with this problem I'm gonna go ahead and draw it. I try to draw the problem to sort of visualize what is happening. So I'm gonna draw my Y axis and my X axis. And we're just gonna draw um line at above the Xbox is out the Y axis for 10. And I'll do one below the X axis at negative 10. And I'm going to go ahead and graph my points. So I have a focus at zero negative 10. So I'll go ahead and do a dot at that position And I have the direct tricks at Y. is equal to 10. So I'm gonna go ahead and do a dotted line where Why is equal to 10 And I'll label my daughter in line Y. is equal to 10. And I know that that's my direct tricks. So recall that problem is open away from the direct tricks towards the focus. So my proble is going to be opening downwards and it could be here or here or here. So we can determine that as we work through this problem. But because I see that the access of cemetery is along the Y axis. I know that my problem will follow the standard form of X minus H squared is equal to four P. Times why minus K. So notice how I have these three variables H. P. And K. That I need to plug in to form my equation. Well we can figure out a church and K. Because the vertex of the parabola is H. And K. So if we figure out the vertex, I figure out two of the variables that I need to solve my equation. So the vertex falls in the halfway point between my direct tricks and my focus. So I need to find the halfway point between these two distances. And if we look at the graph you can see the the direct tricks is at positive 10 and my focus is at negative 10. So naturally halfway would be at zero. But we can also solve for this mathematically if I add them Plus -10 divided by two. The top cancels out to zero divided by two, which leaves me with just zero. So I know that my vertex in this case The x value is at zero. Um the x value is at zero because if we look at the focus the focus is at 0 -10. And the focus is always on the access of cemetery. Of the proble so it will have the same X. Value as my vertex if it's opening up or down. So I have zero and we just solved the Y value equidistant between the direct tricks and focus Is at zero. So my coordinates from my vertex are 00. And I can go ahead and plug those into my standard equation. So my standard equation becomes x minus H. Which is zero squared is equal to four p times y minus K which is also zero. So if I simplify that I have X squared is equal to four P times why? And now I only need to find the value of P. And the value of P is the distance between the vertex and the focus. So it's this little margin here And I'm going to go ahead and update the image of my problem. I know that it opens downward with a vertex at 00 and now I'm looking for P which is this value here between the vertex and the focus. So you can see that the distance between these Is going to be negative 10 from the graph because we're going from zero to negative 10. But we can also solve for this mathematically if we start at The Vertex which has a wide value of zero and then and B. Y. Value of the focus which is negative 10. So minus 10 I got negative 10. Soapy Is equal to negative 10. And then I can also plug that into my standard equation. So now have x squared is equal to four times negative 10. My p value times Y. So this becomes X squared is equal to four times negative 10 is negative 40 Y. So this becomes my final equation. And you can see that this aligns with answer choice. See so hopefully this video helped and see you next time.

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In Exercises 17–30, find the standard form of the equation of each parabola satisfying the given conditions.Focus: (0, - 25); Directrix: y = 25

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