In Exercises 25–36, find the standard form of the equation of eac...

Question

In Exercises 25–36, find the standard form of the equation of each ellipse satisfying the given conditions. Major axis vertical with length 10; length of minor axis = 4; center: (-2, 3)

Accepted Answer

Hello everyone in this video. We're going to be looking at this practice problem where we want to find the equation of the ellipse with the properties that has a major access, vertical length of 14. Length of the minor axis is six and the center is a negative 12. So in order to find the equation for this ellipse, let's recall the standard form of an ellipse or the standard form equation of an ellipse. So the standard form equation is x minus H squared divided by a squared plus X minus or not. X. Again but y minus K squared divided by b squared is equal to one. And for this recall that the center of the ellipse has the coordinates H. K. So we're gonna use this sort of standard form to figure out what our equation is for our ellipse. And a small modification is that because the major access, the major axis is along the vertical link, we need to switch the A squared and the B squared. Because the major access is associated with vertical, we're going to associate it with the Y variable. So it needs to be under the Y variable. In this part we're in this second fraction. So the standard form that we're gonna use is actually x minus H squared divided by b squared plus y minus k squared divided by a squared is equal to one. And in the proper problem prompt were already given the center of negative one and two. Which means that for us, H is equal to negative one and K. Is equal to two. So I can go ahead and write that into my standard form. So I'll have x minus negative one squared divided by B squared plus y minus two squared divided by A squared is equal to one. If I simplify that I have x minus negative one. So plus one squared divided by B squared plus why -2 Squared divided by a squared is equal to one. And now I need to find A. And B. And recall that A. Is the distance from the center to the outer most point on the major axis. So for a I'm just gonna say major access and B. Is the outer moose or the link of the outermost point on the minor axis from the center and were given the length of the Major axis. Were given that it is 14 And the minor axis is six. So if we draw an ellipse the major access is this vertical length here and they're telling us that the length of that is and I'll do the minor axis in red. This minor axis that goes across is six and for A. And B we care about going from the center to those outer most points. So for A we need the value from the center out to this point. So we only care about this part and if you notice that actually represents half of the total length. So for both of these A. or two, a. is equal to the total length of the major axis which is and to be is equal to the total length of the minor axis which is six. Which means that if I saw for A I'll divide 14 by two so A. Is equal to seven. And if I solve for B I'll divide both all divide six by two which means B. Is equal to three. So I can plug those into my standard equation. And if I do that I have X plus one squared divided by B squared. And I'll plug in the value of B which is three plus y minus two squared divided by a squared which is seven, It's equal to one. And if I simplify those I have X plus one squared divided by three times three which is nine plus y minus two squared divided by seven squared and seven times seven is 49 is equal to one. So this becomes my final equation for my lips. And if we look at the answer choices you can see that answer choice A aligns with our answer or what we found. So hopefully this video helped and see you next time

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In Exercises 25–36, find the standard form of the equation of each ellipse satisfying the given conditions. Major axis vertical with length 10; length of minor axis = 4; center: (-2, 3)

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