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: (- 3, 4); Directrix: y = 2

Accepted Answer

Hello, everyone in this video, we're going to be looking at this practice problem where we're going to be writing the equation of the parabola given that it has a focus of negative one and five and direct tricks of Y equals three. So in this case recall that the direct tricks is the line on the opposite side in which the parabola opens up. So if I were to draw a graph, say I have my y axis and my X axis, my direct tricks is Y equals three, which means that the direct tricks lies somewhere above my X axis. And what that's telling me is that the problem will even be opening up towards the bottom or towards the top, which means that our axis of cemetery will be along the y axis. And if the access of cemetery is on the y axis or if my problem is opening up or down there is a standard equation for those sort of problems and they follow the form where we have x minus h squared is equal to four p times y minus k. So we know that we're looking for this sort of problem that is opening up or down. And those sort of parabolas follow this form here. So now I need to figure out what H. Okay and pr and in this case h and k are the vertex of my problem. So say we go back to this imaginary picture the vertex, is that point right at either the top or the bottom of the problem and from the focus and direct tricks, the vertex is the halfway point. So say I'm going to add a slash to my graphs here, I'm going to see that slash is the focus and I'm going to see that we already have the direct tricks on here where y is equal to three. So the halfway point between the direct tricks and the focus is the vertex. So we're gonna use that to figure out the values of the vertex. So you can see at the vertex would have the same coordinates as my focus or x coordinates as my focus. And in this case the x coordinates of the focus is negative one, so I'll have negative one. And because I said that the vertex is halfway between the direct tricks and the focus, I can say that five plus three Divided by two would give me the correct why Value for my vertex. So notice how five is the y value of my focus And three is the value of my direct tricks. So I just want to find the middle ground between those two, so I'm just going to add them together and divide them by two. So this coordinate becomes my vertex and if I solve for this, I get negative one and five plus 3 is eight Divided by two, is 4. So now I know that my vertex is negative one and 4 and negative one in four are my H and key four. My standard form equation. So now I can plug those in, So I have X -H. Which is negative. one Squared is equal to four p. Time is Y -K which is four. And you can see that we've plugged in all the values except for P. Here. So now you have to worry about what P. Is. And from here recall that P is the distance from the focus to the vertex. So now P. Is this value here? If you see in blue it's going from the vertex to the focus. That's the only value which means that P starts at negative one five and it goes to the vertex which we determined to be negative 14. And from here you can see that the value would be five -4 or one So P is equal to one and notice that I got that because we have the five and the four so five minus four is equal to one which means that P. Is equal to one. So if I write rewrite my standard form equation I'll have x negative times negative is positive plus one squared is equal to four times one Times Y -4. And if I simplify this I'll have X plus one squared is equal to four times y minus four. And that would be my final answer. So if I look at the answer choices you can see that that aligns with answer choice A. So hopefully this video helped and see you next time

Struggling with College Algebra?

In Exercises 17–30, find the standard form of the equation of each parabola satisfying the given conditions.Focus: (- 3, 4); Directrix: y = 2

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