Conics from equations
Algebra ||: Conic Sections 39-42, Parabola, Hyperbolas, etc.
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- We're on problem 39.
- And it says which ordered pairs the vertex of f of x is
- equal to x squared plus 6x plus 5?
- Now, finding vertexes of parabolas, there's a formula,
- and negative b over 2a, but you might forget that when you
- turn 30 years old.
- The whole goal of this isn't just to pass an algebra exam.
- It's to learn algebra.
- Then when you learn calculus, you'll learn actually a much
- easier way to figure out minimum and maximum points.
- But there's actually an intuitive way of doing it in
- algebra as well.
- What we have to do is just write this in that form.
- If we could write this function in the form like
- this, f of x is equal to A, times x plus or minus
- something, I don't know, b, squared, plus C.
- Then we know the vertex happens when x is equal to b.
- Because if you look at this form of a parabola-- and you
- can write any of these in this form-- then you say OK, this
- term right here is always going to be positive.
- So the minimum point of this parabola is when this term is
- equal to 0.
- Then the minimum point of this term happens when x is equal
- to b, when this term is equal to 0.
- So the vertex becomes xb comma C.
- So let's do that with this.
- You could memorize it.
- It ends up being minus b over 2A.
- But let's just actually solve it, because I want you to have
- the intuition.
- f of x is equal to x squared plus 6x, plus 5.
- I wrote a gap there for a reason, because we're
- essentially going to complete a square.
- We want to turn this x squared plus 6x, plus something, into
- a perfect square.
- Hopefully you have some practice
- completing the square.
- What number is 1/2 of 6?
- Because when we square it, if we do x plus a squared, that
- equals what? x squared plus 2ax plus a squared.
- So if we map that to that, x squared is x squared.
- 2ax is equal to 6.
- So a would be equal to 3.
- And if a is equal to 3, a squared would be equal to 9.
- You can't just add a 9 to one side of the equation.
- You either have to do it to both sides, or you have to do
- something to offset that.
- So if we added a 9 here, let's just subtract a 9 as well.
- So then we get x squared plus 6x, plus 9, plus 5, minus 9.
- So this right here, this part of it, that's equal to x plus
- 3, squared.
- And then this part, 5 minus 9, minus 4.
- There, we got it into our format that we needed.
- So when does this function achieve a minimum point?
- Well, it achieves a minimum point when this part is equal
- to 0, or when x is equal to minus 3.
- When x is equal to minus 3, this is 0.
- So f of x is going to be equal to minus 4.
- So that's the vertex, and that's choice A.
- You could have done minus b over 2A.
- You'd get minus 12 over 2-- no minus 6 over 2,
- which is minus 3.
- That's something you could memorize.
- But it's better, I think, to have the intuition.
- Problem 40.
- The graph of x over 2, squared, minus y over 2,
- squared equals 1 is a hyperbola.
- That makes sense.
- This is y over 3. x over 2, squared, minus y over 3,
- squared is equal to 1.
- If this is a plus, we would have an ellipse.
- Fair enough.
- Which set of equations represents the asymptotes of
- the hyperbola's graph?
- Once again, there's some equations that people memorize
- and formulas people memorize, but you can actually have an
- intuition on this.
- So let's think about what happens as x approaches
- positive or negative infinity.
- Because that's what we care about as an asymptote, right?
- What does the graph approach when we get to really, really
- large values of x, and it turns out, really
- large values of y?
- So let's just rearrange this a little bit.
- Let's put y on the other side.
- Let's square this.
- So this is x squared over 4.
- Let me keep it actually in that format.
- Let's keep that as x over 2, squared is equal to 1, plus y
- over 3, squared.
- Let's subtract 1 from both sides.
- So you get x over 2, squared, minus 1 is equal
- to y over 3, squared.
- And now, let's multiply both sides by 3 squared.
- Let's multiply both sides by 9.
- Let's do that.
- Actually, to simplify this, let me rewrite this.
- I wanted to keep in that form.
- That's the same thing as x squared over 4, minus 1, is
- equal to y squared over 9.
- Multiply both sides by 9.
- I'm going to switch the sides so we get y here.
- So you get y squared is equal to 9 over 4 x
- squared, minus 9.
- All right, and that's about as far as we can get just with
- straight up algebra.
- When we think about asymptotes, those are points
- that the graphs will never reach but it'll get close to.
- So we have to think about what happens with x gets really
- large and correspondingly y gets really large.
- Well, as x gets really large, so as x approaches infinity--
- and you'll learn this is precalculus and calculus,
- these are limits, but it's really a very intuitive
- concept-- as x gets really large, this minus 9 doesn't
- matter much.
- This term starts to dominate it.
- These terms matter a lot more than this minus 9.
- So as x approaches infinity, y squared is approximately equal
- to 9/4 x squared.
- Now we could take the square root of both sides of this.
- So you get y is approximately equal to plus or minus 3/2 x.
- Those are the actual asymptotes, y equals 3/2 x and
- y is equal to minus 3/2 x.
- That's actually choice A.
- I won't go into the actual instruction of how you would
- think about graphing.
- Actually, why not?
- Why don't we just do that.
- We know what the asymptotes are.
- Let's actually think about how you would graph this.
- So if you were to draw the-- it's all about learning, not
- just about assessing-- that's the x- and y-axis.
- And they tell us their asymptotes, 3/2 x
- and minus 3/2 x.
- So 3/2 x, the slope of 3/2 is 1.5.
- So it's a slope that will look something like-- let me get
- the line tool out-- so x might look like, so plus 3/2 x would
- look something like that.
- Minus 3/2 x would be something like that.
- Now we have to figure out does this hyperbola go from there
- and there, or does it go from here and here?
- There were just say, OK, when y is equal to 0 we get x over
- 2 is equal to plus or minus 1.
- You actually get x would be there.
- So you know that this thing has x-intercepts.
- You can set y is equal to 0.
- You can't set x is equal to 0.
- If you get x is equal to 0, then you get minus y squared
- is equal to something.
- You get y squared is equal to negative 1.
- You get some imaginary solution.
- So you know that it has x-intercepts.
- You know that these are the asymptotes.
- So if you want to draw the graph, it would look
- something like that.
- I know that's not a great drawing, but hopefully it
- serves the purpose.
- Anyway, the choice was A.
- And hopefully you learned a little bit about asymptotes.
- Next problem, 41.
- Which of the following represents a hyperbola?
- Let me copy those.
- No, which of the following represents a parabola, they
- want to know.
- That's too small.
- I don't think you can see that.
- I'll get a bigger version of it.
- All right, that's the bigger version.
- There you go.
- Which of these represents a parabola?
- Well, we could just go one by one.
- This is a circle.
- That's a circle.
- Then you could graph if you like.
- This is an ellipse, where you just kind of weight the y's
- and the x squareds.
- This is an ellipse.
- If this was a negative, this would be a hyperbola, ellipse.
- This one is a hyperbola because you have
- this negative here.
- This one might not look like a traditional parabola,
- but it really is.
- Because if I were to divide both sides of this by 4p, I'd
- get x is equal to 1 over 4p times y squared.
- So it's a sideways parabola.
- It looks something like this.
- If that's the x- and y-axis, the graph will look
- something like this.
- It's going to be a sideways parabola.
- But it's a parabola nonetheless.
- So the choice is C.
- We have time for at least one more, problem 42.
- They give the following equation.
- Let me paste it here.
- All right, and they say what is the standard form of the
- equation of the conic section above?
- And essentially they want us to complete the square with
- the x's and the y's so we get that x minus something
- squared, over some m plus or minus y, minus something
- squared, over some constant.
- So let's do that.
- So let's group the x and the y terms. So you get 4x squared,
- minus 16x, plus something.
- That's how we're going to complete the square.
- Then you get plus minus 5y squared.
- Let me just minus, and I'll do plus 5y squared.
- We have a minus here, so it becomes plus 30y.
- And then out here you have a minus 9 is equal to 0.
- Just to simplify it, let's factor a 4 out of this part.
- So you get 4 times x squared, minus 4x, plus something,
- minus-- let's factor a 5 out-- 5, times y squared, plus 6y,
- probably plus something-- and let's add the 9 to both
- sides-- is equal to 9.
- All right, so let's think about it.
- If we wanted to complete the square here, what do we put
- here to make this a perfect square?
- If this was a plus 4, then this would
- be x minus 2 squared.
- So we're adding a 4 here.
- But we're really adding a 4 times a 4.
- We're really adding a 16 if you think about it.
- If I were to redistribute this, I just added 16.
- So if I add 16 to one side of the equation, I have to add 16
- to the other side of the equation.
- So I'll add 16 there.
- To make this a perfect square, let's see, 1/2 of 6 is 3, so 3
- squared is 9.
- But I'm really adding a 5 times a 9.
- I could redistribute the 5.
- I really added a 45.
- Oh, no, no, no.
- I added a minus 5 times 9.
- I caught myself.
- I added a minus 5 times 9.
- So I'm really subtracting 45 from that side.
- So I have to subtract 45 from that side.
- And so this simplifies to 4 times x minus 2 squared--
- right, that's x minus two squared-- minus 5 times-- this
- is y plus 3 squared-- is equal to-- let's see, 9 plus 16 is
- 25, 25 minus 45.
- Let's see, have I made a mistake someplace?
- 25 minus 45, no, that's right.
- 25 minus 45 is minus 20.
- And now we can divide both sides by minus 20, if I did
- this correctly.
- We'll divide both sides by minus 20.
- So you get x minus 2 squared.
- What's 4 divided by minus 20?
- It's minus 1/5.
- Let me just do that.
- So 4 over minus 20, minus 5, times y plus 3 squared, over
- minus 20 is equal to 1.
- So this is equal to, let's see this becomes a positive, and
- this one becomes a negative.
- So let's put this one first. So 5 over 20 is the same thing
- as y plus 3 squared, over 4-- right, that's this
- one-- plus, no, sorry.
- We have a minus.
- Minus-- and this becomes 1/5-- x minus 2 squared, over 5.
- That is equal to 1.
- We haven't made a careless mistake.
- That is choice B.
- See you in the the next video.
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At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?
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