# Which is the directrix of a parabola with equation x^2=4y brainly. SOLUTION: For the parabola find the vertex,focus and directrix a) x^2 + 6x + 4y + 5 = 0 b) y^2 + 6y 2019-03-29

Which is the directrix of a parabola with equation x^2=4y brainly Rating: 5,1/10 1652 reviews

## calculus This is the same thing as 2yb minus 2yk, which is the same thing, actually let me just write that down. First of all, we will divide both sides of our given equation by 4. And I'm starting to run into my graph, so let me give myself a little bit more real estate over here. . That means that the distance to the directrix, which I'm drawing here in blue, has to be the same as the distance to the focus, which I am drawing in magenta, and when we take the distance to the directrix, we literally just drop a perpendicular, I guess you could say say, that is, that's going to be the shortest distance to that line, but the distance to the focus, well we see that's at a bit of an angle, and we might have to use the distance formula, which is really just the Pythagorean Theorem. So, what's this blue distance? So on the left-hand side, what am I going to have? Examples are shown below, defining a and creating its in this manner. We will now use this information to derive an of this parabola.

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## Write the equation for a parabola with a focus at (1, This can be rewritten as. Substituting the value of a in above formula, we will get: We know that directrix of parabola is. The actual bulb is placed at the focus. Everyone in this room is now dumber for having listened to it. So it's going to be y minus k.

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## SOLUTION: For the parabola find the vertex,focus and directrix a) x^2 + 6x + 4y + 5 = 0 b) y^2 + 6y Given a focus at a point a,b , and a directrix at y equals k, we now know what the formula of the parabola is actually going to be. The of a can be created using a of distances from the focus and from a line, called the directrix, to the graph. The is clearly -1, -5. Step 2: of the of a is given by where h, k are the of the vertex. Well let's get the k squared on this side, so let's subtract k squared from both sides, so, subtract k squared from both sides, so that's gonna get rid of it on the left-hand side, and now let's add 2yb to both sides, so we have all the y's on the left-hand side, so, plus 2yb, that's gonna give us a 2yb on the left-hand side, plus 2yb.

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## Which is the directrix of a parabola with equation X^2=8y You can factor out a 2y, and it's gonna be 2y times, b minus k. So if you square both sides, on the left-hand side, you're gonna get y minus k, squared is equal to x minus a, squared, plus y minus b, squared. Logans2, what you've just said is one of the most insanely idiotic things I have ever heard. Now we have to be careful. Now, on our right-hand side, I promised you a little bit of hairy algebra, so hopefully you see that I'm delivering on that promise. A of this quadratic is shown below.

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## The equation of the directrix of the parabola y^2+4y+4x+2=0 is So that's this distance right over here, and by the definition of a parabola, in order for x,y to be sitting on the parabola, that distance needs to be the same as the distance from x,y to a,b , to the focus. Now, I said all I want is a y on the left-hand side, so let's divide everything by two times, b minus k. Step 1: Find the by. Step 4: The directrix is located below the at. So what's that going to be? Note that the could be rewritten as where d is the focal distance of 1.

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## Determine the vertex, focus, directrix, and axis of symmetry of the parabola with the given equations: It's gonna be our change in x, so, x minus a, squared, plus the change in y, y minus b, squared, and the square root of that whole thing, the square root of all of that business. This general form of the for a tells us that if , then the focal distance d is because. Step 3: Since the of the opens upward from the vertex, the focus is located at which is above the vertex. The combined distances from these foci is used to create an of the and. Well, we just apply the distance formula, or really, just the Pythagorean Theorem. So, for example, if I had a focus at the point, I don't know, let's say the point 1,2 , and I had a directrix at y is equal to, I don't know, let's make it y is equal to -1, what would the equation of this parabola be? I can always check my graph, if I'm not sure about this.

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## which is the directrix of parabola with equation x^2=4y So what we really want is the absolute value of this, or, we could square it, and then we could take the square root, the principle root, which would be equivalent to taking the absolute value of y minus k. Thus the of the directrix is. This puts the focus 3 units to the left of the vertex. Well, that simplified things a little bit, and now I can, let's see what I can do. At no point in your rambling, incoherent response were you even close to anything that could be considered a rational thought. So let's take a arbitrary point on the parabola. Answer: Step-by-step explanation: We have been given an equation of parabola.

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## Which is the directrix of a parabola with equation X^2=8y Well, that's just gonna be our change in y. So what is this going to be equal to? It's going to be this y, minus k. So we could write this as 2 times, b minus k, y if you factor out a 2 and a y, so that's the left-hand side, so that's that piece right over there. Why we do this: One application of a involves lighting. Now, we will compare our equation with vertex form of parabola: , where, h,k represents vertex of parabola. The way I've just drawn it, yes, y is greater than k, so this is going to give us a positive value, and you need a non-negative value if you're talking about distances, but you can definitely have a parabola where the y-coordinate of the focus is lower than the y-coordinate of the directrix, in which case this would be negative.

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## find the equation of the parabola with focus at (0, Now, this right over here is an equation of a parabola. Find the vertex, the focus, and the directrix. We are asked to find the directrix of our given parabola. And I'm actually gonna divide this whole thing by two times, b minus k. Now let's see if we can simplify things. Now, we will find distance of focus from vertex of parabola using formula. Now, obviously on the left-hand side, this all cancels out, you're left with just a y, and then it's going to be y equals, y is equal to one over, two times, b minus k, and notice, b minus k is the difference between the y-coordinate of the focus, and the y-coordinate, I guess you could say, of the line, y equals k, so it's one over, two times that, times x minus a, squared.

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## calculus This distance has to be the same as that distance. Derivation: Since the focus is 0, 1 and a general on the directrix can be given by x, -1 , the would give us. We can see that the value of a is , and. Let's Practice: We will now examine two examples relating the equation, vertex, and focal distance. Examination of the tells us that the of this is the origin. The focus of a has a property which is important is constructing lamps and lighting. On the right-hand side, you have x minus a, squared, and then, let's see, these characters cancel out, and you're left with b squared minus k squared, so these two are gonna be b squared minus k squared, plus b squared minus k squared.

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