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, 2); Directrix: x = - 1

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 a problem standard form with the conditions that the focus is at 61 and the direct tricks is X. Is equal to negative four. So in this case before we even get started on the solving, I'm going to try to visualize this. So I'm going to try to draw the information that they gave me. So I'm gonna draw the Y axis and the X axis and without adding any check marks or labor ling any numbers. I'm going to say that my direct tricks is X. Is equal to negative for which means that it would lie somewhere left of my Y axis. So that's where I'm going to put it. I'm gonna make it the started line here and say X. Is equal to negative four. So that's my direct tricks. And remember that the direct tricks is on the opposite side of the problem opening. So I'm gonna keep that in mind. And then My focus is at 61 which means it's somewhere to the right of my why access. So that's where I'll put it. I'll say that my focus is somewhere here. Call that six one. So because my direct tricks is giving me next value, I know that my axis of cemetery is somewhere on the X axis which means that the problem will either open to the right or it will open to the left. So these are the two problems that I have possible right now. So now that I've drawn this picture, I'm going to go ahead and get started at the problem. So because my problem will either open to the left or to the right, I know that it will follow the standard equation of why minus K Squared is equal to four p x minus H. So now I need to worry about finding these values K. P. And H. And we can start by determining the vertex which is H. Key. And the vertex is this point here that is equal distant from the direct tricks and the focus. So I think from this picture it's kind of hard to tell that It's equal distant but essentially that Vertex is the same distance from X is equal to negative to The .61 or the focus. So we know that it falls along the same line as the focus, which means that the y value will be the same as the focus. So the y value will be one. So I know that K will be equal to one. You can see here since I know that why value for my vertex. So I'm just gonna say K is equal to one but I need to solve for my X value of my vertex and I know that it's halfway between my focus and direct tricks. So if I add the X values of my direct tricks and focus and divide that by two, I can determine that value. So I can do negative for plus six divided by two. And that will be the X value of the vertex. So if I do that negative four plus six is equal to positive two divided by two is equal to one. So my vertex is actually at one one which means that H is also equal to one. And also that my drawing is slightly off, it would be closer looking to this with the vertex still being at this point here. So notice how we can sort of update the image of the problem as we solve different things. And now that I have K and H I can go ahead and rewrite my standard equation as y minus K which is one squared is equal to four P times x minus H which is one. And now I only need to solve for P and recall that P is the distance from the focus, the focus the dot here to the vertex. So p is this distance. So notice we can determine that because they are on the same access, they are both at This y value which is one. So the only thing that varies is there X values. So we just need to figure out how apart how far apart there X values are and that will give us the value of P. So I know that my focus has X value of six and my vertex has an X value of one. So if I do six minus one, I get five. So that's the value of P. Or the distance between the vertex and the focus. So I can plug that back into my equations. Now I have y minus one squared is equal to four times five times x minus one. If I do this multiplication, I'm left with y minus one squared is equal to times x minus one. Now this would become my final answer or the standard form of The problem that has a focus of and a director X of X -1. And notice that by working out this problem, we determined that the problem opened up to the right, and also our final answer becomes B. 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: (3, 2); Directrix: x = - 1

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