Algebra (all content)
- Representing a line tangent to a hyperbola
- Common tangent of circle & hyperbola (1 of 5)
- Common tangent of circle & hyperbola (2 of 5)
- Common tangent of circle & hyperbola (3 of 5)
- Common tangent of circle & hyperbola (4 of 5)
- Common tangent of circle & hyperbola (5 of 5)
- Intersection of circle & hyperbola
2010 IIT JEE Paper 1 Problem 45 Circle Hyperbola Common Tangent Part 2. Created by Sal Khan.
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- I'm watching this video in the Algebra: Conic sections topic. What does IIT JEE mean?(5 votes)
- How does the IIT JEE compare to other Standardized tests in America, like the SAT or ACT?(0 votes)
- IIT JEE is a test that requires a lot of intelligence, practice, effort and time-management. SAT is just for kids. SAT requires taking excellent English classes in school, and practicing time-management. The SAT and IIT JEE are not exams that can even remotely be compared.
I took the IIT JEE in 2011 and got an All India Rank of 4249 (out of 5,00,000 candidates).
I also took the SAT with a week's preparation (only took practice tests at home a week before the exam) and scored 2100/2400. I haven't taken the SAT Subject Tests but seeing the sample questions, I feel even the Subject Tests are not anywhere close to the IIT JEE.
But I won't agree with Rishi (above); IIT JEE is NOT of the same standard as the IPhO or IChO. The Olympiads are tougher.(16 votes)
- what does IIT JEE mean?(3 votes)
- At8:30, why isn't it 8(sqrt(m² + 64)) instead of 8(sqrt(m² + 1)) ? It looks like it should go from sqrt(8²m^2 + 4•16) to sqrt(m² + 64) • sqrt(8²) to 8(sqrt(m² + 64)). Thanks!(1 vote)
- You need to factor 64 out of both terms.
sqrt(8²m^2 + 4•16) = sqrt(64m^2 + 64) = sqrt(64(m^2+1)) = sqrt(64)sqrt(m^2+1). Remember that square roots separate across multiplication, not addition.(5 votes)
- While Sal is doing the quadratic equation he assumes that b=8m, not 8mb, but in the quadratic from which he is deriving the quadratic equation the b term is 8mb. Why the omission of the b term?
Also, shouldn't the c term be -16, why does Sal write a positive 16?(2 votes)
- Sal Probably shouldn't have defined the y-intercept as b as the b in the quadratic equation is only the coefficient of b, and that is 8m. C is -16, but it got multiplied by -4, so it becomes positive.(3 votes)
- First of all, thank you for this video.
Big picture: I am confused about using math "tricks". In particular the "trick" Mr. Khan uses around1:20of substituting an entire general equation into a specific equation.
Early in the video (around1:20) Sal, put a generic equation of a standard line (y-mx+b) into a SPECIFIC equation for a circle. To me this seems like math wizardry and I am VERY curious about receiving an answer on this. It appears to me you could throw any number of equations and play with them algebraically, how do you know which equation to use? and where to put it?
When Mr. Khan does the tricky "add 1 to both sides" (and 1 can be anything divided by itself).. I can follow that very well. But this is blowing my mind.
When can you use this "put a standard equation into a specific equation" trick?
As a side note, Mr. Khan also did some "put a standard equation into a specific equation" trick/wizardy in the previous video ("Common tangent of a circle & hyperbola (1 of 5)). I ignored it because he said something like "if you're taking this specific test, just remember this rule for time purposes". So I ignored it.
But this is the 2nd time he has used it. And I really don't understand when, why, and how to use this trick/wizardy/tool.
Summary: When, why and how can you substitute "standard" (y=mx+b) equations in specific equations for an object (circle, hyperbola, etc)?(2 votes)
- the answer to when can we substitute standard equation of a line in specific
when you will have to find an equation of either the tangent to the given object or the intersection points when suppose a line having general equation y=mx+c either intersects it or is a tangent to it
why and how:
it's just like solving simple quadratic equations !! whenever you are supposed to find intersection points then you will always have to solve ''simultaneously'' the given equations as sal did in the video he just simultaneously solved by substituting the equation of line i.e with degree 1 into another equation of degree 2
so the point is that in general to find the equation of tangent that is line (y=mx+c) or intersection points you have to use the standard equation and conics equation and simultaneously solve them
hope you get the point
Thank You.. !(2 votes)
- I understand why we set b^2-4ac to equal zero. What I'm wondering is what happens when we do that? Don't we mess up the algebra when something that might had not equaled zero now equals zero?(1 vote)
- The whole equation becomes -b/2a, but that is allowed in algebra, so we're not messing anything up.(3 votes)
- Exercises? Are there any exercises?(1 vote)
Now that we have a visual sense of what this common tangent with a positive slope would look like, let's see if we get some constraints on it, especially constraints on its slope and y-intercept. So this line that I drew in the last video here in pink-- it would have the form y is equal to mx plus b. It's a line where m is the slope and b is the y-intercept. Now, let's think about what constraints there would have to be on m and b if this is tangent to the circle. And you might be tempted to break out some calculus and figure out the slope at any point alongside a circle, but there's an easier way to do it. You just have to realize that if a line is tangent to a circle, it will only intersect that circle at one point. Let me show you what I'm talking about. So this is the line. What I want to do is figure out where this equation and the equation of the circle intersect. That's we'll focus on this video. And then, we'll do the same thing for the hyperbola. So we have y is equal to mx plus b is the equation of the line. The circle-- they give us the equation up here. x squared plus y squared minus 8x is equal to 0. So the circle is x squared plus y squared minus 8x is equal to 0. So what we can do is we can substitute this expression over here-- we can substitute this in for y. And then we can figure out what are the constraints on m and b so that we only have one solution to the intersection where we only intersect at one point. So to do that-- actually, let's substitute for y squared. So if y is equal to this, let's square that. We'll get y squared. I'm just squaring the expression for the line. y squared is going to equal m squared x squared plus 2mbx plus b squared. All I did is I squared this expression here. And I did that so that now we can substitute this whole thing right in here for y squared. And the expression for the x point for our intersection is going to be x squared plus all of this business. That's the y squared. Plus m squared x squared plus 2mbx plus b squared minus 8x is equal to 0. And if we wanted to write this as a quadratic in terms of x, this would be-- so our x squared terms are these two terms. We could write this as-- let's see. Let's write this as m squared plus 1 times x squared, right? m squared times 1 times x squared. And then our x terms are this one and this one. So then we have plus 2mb minus 8 times x. And then, we just have this b term-- this b squared, the constant term right over here. And I'll do that in orange. So plus b squared is equal to 0. So if we knew m and b, if we knew the equation of this line, this would just be a straight up quadratic. You could use the quadratic formula to figure out the x values-- where they intersect. Now, what's neat about this is we know that they only have to intersect in one point. Remember, the quadratic formula-- negative b plus or minus the square root of b squared minus 4ac. All of that over 2a. And don't get this b confused with the y-intercept b. This is just from the quadratic formula. That's the quadratic formula over there. This will only have one solution if this over here is equal to 0. Because you're just adding and subtracting 0, so you're only going to get one solution. So when a line is tangent to a circle, it can only intersect in one point. Or another way to think about it-- this will only have one solution. If a line intersects any other type of non-tangent line, a non-tangent line would do something like that. It would either have two solutions, in which case, this is a positive value, or it won't intersect at all. And then, this will have no solutions, which means that b squared minus 4ac would be a negative number. So we know that this is a tangent line. So we only have one solution, or b squared minus 4ac is equal to 0. So what's b squared minus 4ac over here? Well, this is our b when we think in terms of the quadratic formula. And remember, don't get that b confused with the b of the y-intercept. I'm just thinking of the quadratic formula here. So let's do this. Let's take this squared. So I'm just going to rewrite this expression and set it equal to 0, because we know there's only one solution. So we have 2mb minus 8 squared. And then you have minus 4 times a, which is m squared plus 1, times c, is b squared. And so this is going to have to be equal to 0 if this is truly a tangent line. So let's see any type of interesting things that we can get out here. If we can express b as a function of m, that's a good place to start. So let's try to do that. So let's see. If we expand this out, this becomes 4m squared. Let me do that in the same blue so you know what I'm expanding out. This part over here becomes 4m squared b squared minus 2 times 8 is-- 2 times negative 8 is negative 16. Multiply that times 2, so it's negative 32mb-- I'm just squaring this over here-- plus 64. So that is that term over there, expanded, minus 4 times-- well, I could just expand everything out. Minus 4 times m squared b squared minus 4 times 1 times b squared is all going to be equal to 0. And then, lucky for us, some of these terms cancel out. 4mb squared. Negative 4mb squared. And let's see. We could actually divide both sides of this equation by 4, and we get negative 8mb plus-- we're dividing everything by 4-- so plus 16 minus b squared is equal to 0. And now, we can solve for b in terms of m using the quadratic formula again. So now we would have a constraint, or we would essentially know what our y-intercept is going to be in terms of our slope. And then, we can do that for the hyperbola. And then, we could essentially say, well, it's the same line so the y-intercepts have to be the same. And then, we can solve for the slope. So let's do that. And you'll see that over the next few videos. Let me just write this in a form that we would recognize. This is the same thing. Let me just multiply this equation right here-- both sides by negative 1. So then it would become b squared plus 8mb minus 16 is equal to 0. I just multiplied this by negative 1 and just rearranged the terms. Now let's solve for b in terms of m. So b is going to be equal to negative 8m plus or minus the square root of this term squared. So it's 8 squared m squared minus 4 times a, which is just 1, times c, which is 16-- minus 4 times negative 16. So you could view this as plus 4 times 16. All of that over 2a. Well, a here just 2. All of that over 2. Now, this is going to be equal to negative 8m plus or minus-- now, this is 64. This is 64. So you could factor out the 64 from here, but when you take the square root of it, it's going to be 8 times m squared plus 1, right? If you took the 8 in, you'd have to square it. So it becomes 64. And the 64 times m squared plus 64, which is exactly what you had up there. All of that over 2. And then, we can simplify it. This is equal to negative 4m plus or minus 4 times the square root of m squared plus 1. So this is a possible b, given that the line is tangent to the circle. Now, let's just think about this a little bit. If we add 4, we're definitely going to have-- well, let's think about it for second. If we look at the line up here, the way I drew it-- we want a positive slope. And in order to do that-- the way I drew it-- you have to have a positive y-intercept. Let me just write it this way. This is a positive b, a positive y-intercept. So we want this value. We want to think about the y-intercept that is positive. Now, m is going to be positive. We know from the problem that we're looking for positive slope. So m is positive. So negative 4, this whole term here, is going to be negative. So our only chance of being positive is if we add 4 times this expression right over here. And actually, if you look at it, it will be positive, because this is greater than m squared. So the square root of that's going to be greater than m, so 4 times this is going to be greater than 4m. So if we add, it's going to be positive. So we want to only look at b is equal to negative 4m plus 4 times the square root of m squared plus 1. I'll leave you there for this video. In the next video, we're going to do the exact same thing for the hyperbola realizing that the line will only intersect at one point. And then, since it's the same line, we know that their b's have to be the same. In the next video, we're going to get b is equal to some other function of m. We're going to get that in the next video. And then, we can set them equal to each other and solve for our m. And then, when once you solve for an m, you also have solved for the b. And we'll have our line.