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
How a tangent line relates to a hyperbola. Might be useful for some competitive exams where there isn't time to derive (like we are doing in this video). Created by Sal Khan.
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- In the video Sal says he did conic sections IIT JEE problems in previous videos, but this video is the first in the series. Where can I find the previous videos?(10 votes)
- There are many IIT videos here: https://www.khanacademy.org/test-prep/iit-jee-subject/iit-jee - several of them involve conic sections, so he's presumably referring to those.
(It's weird, those are off in their own little "Test prep" category, instead of being in the "Math" category. Not sure what the purpose is of hiding them over there. Seems like they should be moved here.)(6 votes)
- I'm currently in Calc 2 and we worked on conic sections.....so how is conic sections applicable to Calc 2?(5 votes)
- Conic sections provide a rich zoo of possible curves and scenarios for use in many methods learned in calculus. I don't know which Calculus 2 curriculum you are in, so I cannot give an exact mapping. Some schools offer Calculus over 3 Semesters (or more), while others offer it over 4 Quarters, or a more intensive series in 2 Quarters. Calculus for non Math/Physics Majors is different from calculus for Math and Physics majors. But anyway, across all the topics of Calculus, conics keep popping up: just think of solids of revolution...surface areas and volumes...parametric representation of curves...finding slopes of tangents to whatever conic...rotating axes of curves for conics...parametrics...particle motion...and so on.
Yum. Knowing the various relationships of conic sections gives you avenues into setting up otherwise impossible problems for successful solution.(11 votes)
- How many questions are in the exam? And subsequently how much time?(2 votes)
- These were the JEE Advanced papers:
each paper is for 3 hours.(6 votes)
- It says that the video is, "very optional" so will this be on a precalculus final or not?(4 votes)
- This might sound like a basic question, but what does IIT JEE mean? What does it stand for?(2 votes)
- is this relationship that Sal found in the end "9:30" is true also for a hyperbola opens up and down ??(3 votes)
- At6:00, what if the coefficient of the x^2 term equals zero? Then, m^2-b^2/a^2 = 0 , and it will become a liner equation which it still has only one solution.
I hope my question is clear enough :-)(2 votes)
- Well, then line has an equation y = b/a*x + c or
y = -b/a*x + c. This lines are asymptotes of hyperbola shifted up(down) by c units. They intersect hyperbola in only one point, but they are not tangents. I wonder myself, why this case was missed in the video?(2 votes)
- What would be the tangent line relation for the hyperbola (y^2/a^2) - (x^2/b^2) = 1 ?
Would it be x = my + c where c^2 + b^2 = (a*m)^2 ??
What would be the relation for the hyperbola xy = a constant ?(2 votes)
- Why did sal stop when he had the hyperbola equation equal to y^2? Why didn't he just take the square root of both sides to get y=b/a*x-b?(2 votes)
- Maybe because the square root of [b^2x^2/a^2 - b^2] does not create b/a*x-b
You can not split the square root across terms. It just won't work.
For example: sqrt(16+9) = sqrt(25) = 5
If you use your methoed: sqrt(16+9) = sqrt(16) + sqrt(9) = 4 + 3 = 7. This is an incorrect value.
Hope this helps.(1 vote)
- At5:06, how does Sal arrive at (m^2-b^2/a^2) x^2? What happened to the x^2 term that was part of b^2/a^2?(1 vote)
- It was not lost. Sal didn't write out all the steps. Here's basically what happened...
Sal started with: m^2x^2 - (b^2x^2)/a^2
He has 2 terms that contain a common factor of x^2. He used the distributive property to factor out the x^2. This creates: m^2-b^2/a^2) x^2
Hope this helps.(2 votes)
After going through many of these Indian Institutes of Technology Joint Entrance Exam problems, I realize that there are a lot of problems where they really just expect you to know something. So that's what I'm going to cover in this video, one of those things that they just expect you to know. What we're going to do is come up with the relationship between a conic section-- and a particular hyperbola is what we'll focus on-- and a tangent line. We've done this in a previous problem. But that wasn't in the general case. So let's say we have a left right opening hyperbola. So it'll have the equation, x squared over a squared minus y squared over b squared is going to be equal to 1. And so if I were to draw that hyperbola it would look something like this. That's the x-axis. That's the y-axis. And then it opens to the right. I could draw a better bottom half. It opens to the right. And it opens to the left. And in case you're curious, this point right here, if you set y is equal to 0, this point right here is a comma 0. And this right here is negative a comma 0. So what I want to do is figure out a relationship between these a's and b's and the equation of a tangent line. So let's say I have a tangent line that looks something like this. Let's say it's tangent only at that point right over there. And so it would look like-- let me draw it a little bit better than that-- it would look something like that. And let's say the equation for this tangent line is y is equal to mx, where m is the slope, plus-- instead of saying b for the y-intercept. So normally, we would call the y-intercept b for a line. We've already used the b here in the equation for the hyperbola. So let me just call this c. So the c-- this is a little unconventional. This is going to be the y-intercept. So let's see if we can come up with a relation between the m's, the c's, the a's, and the b's. And we already used it in one of the IIT problems. And I suspect that the next one I'm going to do will also use this. If you all have seen a lot of Khan Academy videos, you know that I always like to prove things from first principles, because in life, you can't just memorize formulas. You won't know where they came from. You'll memorize them wrong. You won't understand what they actually mean. But if you're going to take the IIT JEE exam, I would recommend that you-- I have an appreciation for how little time they give these problems. And if you have to prove from first principles, you won't be able to prove them. So let's just come up with the relation. Let me stop talking without drawing. So let's just see where they intersect. And the whole insight here is that they're only going to intersect in one point. So what I'm going to do here is solve for y squared. So over here, we can multiply both sides of this equation by-- let's multiply both sides by negative b squared. So you get negative b squared over a squared x squared plus y squared-- I multiply by negative b squared-- is equal to negative b squared. And now, let's add this thing to both sides of this equation. And we get y squared is equal to b squared over a squared x squared minus b squared. So I just rewrote the equation for the hyperbola. And let's also write this in terms of y squared. And then we can set them equal to each other. So over here in this greenish yellow-like color, if we square both sides, we get y squared is equal to m squared x squared plus 2 times the product of those both terms. So plus 2 mcx plus c squared. So in order for them to intersect, they both have to be at the same place at some x and y. So we can set this y squared being equal to that y squared, and then try to solve for the x. Obviously, we won't be able to solve for the x, because there's so many variables. We can find a relationship between this a, this b, this m, and this c, so there's only one point of intersection, which by definition, it would have to be at the tangent point. So let's do that. So we have m squared x squared plus 2 mcx plus c squared is equal to b squared over a squared x squared minus b squared. And I've done a very similar exercise to this in a previous IIT JEE video. But here, I just want to focus just on the most general case so that we have something that we can add to our tool kit. So let's write this in terms of a quadratic equation in x. So if we subtract this from both sides, we get m squared. We have m squared minus b squared over a squared x squared. That's that term and that term, right over there. And then my only just first degree x term is right over here. So plus 2mcx-- I'll write it in-- plus 2mcx. And then finally, plus, we have a c squared, and then we're actually going to have a plus another b squared over there. So this is going to be plus c squared plus b squared. Now, in order for this equation to only have one solution-- let me write it down. This is going to be equal to 0. In order for this thing to only have one solution, the discriminant of this quadratic equation-- remember, when you do the quadratic equation-- and these are completely different, so negative b plus or minus the quadratic formula b squared minus 4ac over 2a-- you're only going to have one solution if this thing over here, if the discriminant over there is equal to 0. If b squared minus 4ac is equal to 0, then you only have the solution negative b over 2a. So in this situation, for the tangent line, you can only have one solution, one x that satisfies this equation. So b squared minus 4ac is going to have to equal 0 in the quadratic formula. These are different b's and a's and c's than the ones we're using here. Over here, our b is that right over there, the coefficient on the x term. So that squared is 4m squared c squared. And we're going to subtract from that minus 4 times a. a is all of this business right over here. And just to simplify things, let me make this plus, and then multiply this times a negative sign. And it'll reverse the sign. So I'll put minus 4-- I'll put a plus here-- times, instead of writing, since I have to multiply this by negative 1, it'll now be b squared over a squared minus m squared. And then the c is this term right over here. This right here is c, 4ac. c squared plus b squared. And this thing is going to equal 0 if this line is tangent, if we only have one solution. So the first thing that we can do to simplify this is, we can divide both sides of this equation by 4. And if we do that, this becomes-- I wanted to do that in black-- this becomes a 1. So we can ignore that. And then this becomes a 1. So that simplified our equation a good bit. And now, let's multiply the second part right over here. So we have b squared over a squared times c squared. That is b squared c squared over a squared. And then b squared a squared times b squared-- so plus b to the fourth over a squared. Then you have negative m squared times c squared. So let me do this in another color. So then you have for the m squared, negative m squared c squared. And then you have negative m squared b squared. And of course, this is all going to be equal to 0. And we have this m squared c squared out front over here. And lucky for us, this cancels with this. And what do we have left? Now, every term here is divisible by b squared. So let's divide every term by b squared. So this will just become a 1. This will become b to the second power. And then this will just become a 1. And then let's multiply everything by a squared, just so we get rid of the fractions. So when you multiply everything by a squared, this term right over here becomes a c squared. This term right over here is just a b squared. And then all we have left, and then we have this negative m squared. Remember, we're multiplying by a squared. So minus a squared m squared is equal to 0. Or we could add this to both sides of the equation. We get c squared plus b squared is equal to a squared m squared. And what's really neat about this is, we now have a very simple relationship. If we know the equation of the line right over here, we would then, if we know what m and c are, we then have an interesting relationship for a and b. If we know what a and b are, we have an interesting relationship for the equation of the line. And maybe if we have a few other constraints, we could actually solve for them. But we'll actually take this and use this in the next IIT problem we're going to do.