Algebra

Conic sections: Intro to hyperbolas | Conic sections | Algebra II | Khan Academy

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Let’s see if we can learn a thing or two about Excess cut And away from all conical sectors, perhaps this Some people will confuse you a lot, because it is not Easy to draw as a circle and ellipse We have to use some algebra But I hope that through the show you will be Somewhat relaxed, and you will see that the extra sectors Somehow it is more fun than conical sections A little review, I want to do this to see Similarities between formulas or ideal shape For different conical sectors If we have a circle with a center of 0, and its equation x ^ 2 + y ^ 2 = r ^ 2 And as we saw, this can be written as – and I do This is because I want to show you this The same familiar equation for ellipse If we divide both sides by r ^ 2, we get x ^ 2 ÷ r ^ 2 + y ^ 2 ^ r ^ 2 = 1 This is why this is a circle And again, the circle, all the points located on The circle has the same distance from the center Or in this case, you can say that the primary axis The sub-axis has the same dimension, as it does Difference between the two points The distance is the same from the center So this is the circle This ellipse, but these two numbers They can be different Because the distance from the center can change So x ^ 2 ^ a ^ 2 + y ^ 2 ÷ b ^ 2 = 1 This is the ellipse And now, I’ll skip the parabola, because it is Interesting case, we have reviewed previously So I’m going to dig deeper into it in the future But the hyperboloid formula is close to this There are two ways in which you can write the hyperbola And I will do it It can be written as x ^ 2 ^ a ^ 2 – y ^ 2 ^ b ^ 2 = 1 And notice that the only difference between this equation and that So instead of (a + y) ^ 2, we have a- y So this is one plus cut The other is if the signal is negative The other way If y ^ 2 ^ b ^ 2 – x ^ 2 ÷ a ^ 2 = 1 So now the negative sign in front of the expression x ^ 2 instead From the phrase y ^ 2 What I want to do now is try to find it, how Can we plot each of these equivalent sectors? Perhaps we can use the two methods With the help of books, or if you extract them From the internet, they will give you formulas But I do not prefer these formulas First, because I always forget it And you’ll forget it quickly after the exam You may want to save it if you wish Solve the exam quickly But you’ll forget it The second reason is that you will not only forget it, but you You will feel confused Because it is sometimes used a below x and b Below y, or sometimes you use a below The positive statement and b below the negative phrase If you save, a ÷ b is a slope Converged line and all of this, therefore can be used a and b are wrong So I encourage you to extract it yourself This is what we will do now It won’t take much time So these are two excess herds And what I want to do is that Find the value of y when I have a hyperboloid In this case, if you subtract x ^ 2 from the square image from Both ends, I’ll get – let me change the color – I’ll get -y ^ 2 ^ b ^ 2 You will be here = (1-x) ^ 2 a ^ 2 Then, to see, I want to get rid of this negative signal, and I want to get rid of b ^ 2 So let’s multiply both sides of the equation In -b ^ 2 If you hit the left side x -b ^ 2 The negative and b ^ 2 are omitted, and we remain y ^ 2 = -b ^ 2 Then -b ^ 2 x a = (a + b) ^ 2 ^ a ^ 2 x ^ 2 we are here And we get y = – and I do that Purpose – positive or negative square root, as it can be Positive or negative square root Let’s flip this, to have it Positive statement first b ^ 2 a ^ 2 x ^ 2 – b ^ 2 Now she says, she told us that this will be easy I am solving it And it looks like something complicated But remember, we’re doing this to find converging lines To cut off the excess, to make you feel That we are doing the solution Let me do it here – actually, I want to do Another plus So the hyperbola, if this is the x axis, and this is the y axis They have two similar lines The two closely related lines are these two lines Which the two extra pieces approach So if these two lines are converging – they are always The negative slope of the last – we know this Plus cut, I will quickly clarify any One of them, will look like this, where We are all near endlessly, we will come closer to this line And that line And it would look like this – I didn’t draw it Well; He does not touch the line But it is getting closer and more That is, closer to the converging line It will either resemble this, as it is open Up to left and right Or, the extra pieces will be open up and down And again, as we get closer and closer, and The converging line means that it will be more and more close to one These lines without touching them He will come closer the further we get away And x gets bigger To determine which of these, let’s Think about what will happen when x gets bigger So when x gets bigger So when x approaches an infinite infinite, or x approaches a negative infinite I will say infinitely positive or negative, right? This does not matter, because when we take the negative This becomes square So this number becomes huge as we get closer to Positive and negative infinity You will learn more about this when we develop Boundaries, but I think that’s self-evident This number will become huge This number is fixed It will remain the same The more x is approaching infinitely positive or negative, the more it becomes Too big, too, and y’s value to equal Actually, I think that’s appropriate . Always forget to blog Approximately This means not exactly but almost equal When x approaches infinity, it will be almost Equivalent to, + or – square root of b ^ 2 ^ A ^ 2 x ^ 2 And that’s equal to– now we can take the square root You cannot take the square root if the value is algebraic But you can for this This is equal to + or – b ÷ ax And this tells us, what are the converging lines Which is on the slope of one line and it will be b ÷ ax This will give the positive output b ÷ ax, and the other It would be negative b ÷ ax And I will do this with other examples Ki Make things clear But we still do not know what the shape of converging lines These are the two lines But one is positive b ÷ ax, y = + b ÷ ax Let’s say he’s the first This line here is y = + b ÷ ax And I know you can’t read that Hence the bearish slope may be called y = – b ÷ ax So here are the two lines But we still have to see if the extra cut is open upwards Left and right, or is it open up and down? Here, there are two ways to do this The first, well, that’s an approximation That is, when we close x infinity But as we can see here, when x approaches infinity Then the value will be less than that number Because we subtract a positive number from this We subtract a positive number, and then we take The square root of the total number So we always go for value to be less than The line, especially when we’re on the positive quarter True? For me, I love doing this I think we always – at least in this positive quarter This will be confusing when we go to Other quarters – the value will be just under From the asymptote So we will get close to the bottom And when we know that we are here, you will know that it will be Such and approaching this line Then when it is open right It will also be open to the left Another way to test this is, perhaps, to be more intuitive For you, please find in the original equation Can x or y be equal to 0? Because when it opens right and left So x cannot be equal to 0 We get y = 0, here and here But we don’t get x = 0 Perhaps your teacher will ask you to appoint Points, and then we replace y = 0 You can only look at the original equation In fact, you can look at this equation Can x be equal to 0? If you look at this equation, and if x = 0, all of this The phrase will be deleted, and it remains We have (ab) ^ 2 That is, we take b ^ 2, and we put Negative sign in front of it So this is a negative number Then we take the square root of a negative number So we are not dealing with mock numbers now We cannot get x = 0 But y can be equal to 0, right? So we can put y = 0 and solve it So in this case, let’s actually do this If y = 0, we get 0 = square root For b ^ 2 a ^ 2 x ^ 2 – b ^ 2 So if you square all sides, you’ll get b ^ 2 a ^ 2 x ^ 2 – b ^ 2 I know this is messy Then we get b ^ 2 ^ a ^ 2 x ^ 2 = b ^ 2 You can divide all sides by b ^ 2 You get 1 and 1 Then you can multiply all sides by a ^ 2 You get x ^ 2 = a ^ 2 x = + or – the square root of a So this point is a, 0, and this point – a, 0 Let’s move on to the next issue I feel I do not have enough time Note that when the expression x is positive, then the extra slash will be Open left and right And by deductive analysis when Y is positive, which is this case The cut will be open up and down Let us prove this to ourselves Let’s find y We get y ^ 2 ^ b ^ 2 We will add x ^ 2 ^ a ^ 2 to both sides So we get x ^ 2 ^ a ^ 2 +1 We multiply all sides by b ^ 2 y ^ 2 = b ^ 2 a ^ 2 x ^ 2 + b ^ 2 We have to distribute b ^ 2 We take the square root I will change the color So y = + or – the square root of b ^ 2 ^ A ^ 2 x ^ 2 + b ^ 2 And again we can work The same proof where x approaches = or – Infinity, this equation, this b, this is a constant phrase It will not have much impact We have to take the square root For this phrase It is b ÷ ax, + or – b ÷ ax And once again watch out, these are the two approaching streaks, which are I redrawn them here, this and this But in this case, the output will be slightly larger From this line Slightly larger than this line In the positive quarter, it will be Up here and below here Another way of thinking, in this case, is when The hyperbola is vertical, that is, open up And below, you’ll notice that x can be equal to 0, but y never Will not equal 0 This is also logical Because if we look at the original formula here x can be equal to 0 If x = 0 We can find y But if y = 0, it will not be -x ^ 2 a ^ 2 = 1, and then if we solve This, we get x ^ 2 = – a ^ 2 And we don’t deal with fake numbers, so we can’t Squared number, and we don’t get a negative number Once again, this will be impossible So this is the other key that tells us that it is open up and down Because y in this case is not equal to 0 However, you may have been a little bit confused because I stayed Get b and a In the next presentations I will solve a set of questions He sketched several excess sectors, missing sectors, and Circuits with actual numbers see you soon

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