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## Algebra (all content)

### Course: Algebra (all content) > Unit 17

Lesson 12: Challenging conic section problems (IIT JEE)- 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

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# Representing a line tangent to a 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.

## Want to join the conversation?

- 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:

Paper 1:http://www.jee.iitm.ac.in/JEE2014/docs/2014p1key.pdf

Paper 2:http://www.jee.iitm.ac.in/JEE2014/docs/2014p2key.pdf

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)
- Indian Institutes of Technology- Joint Entrance Exam(4 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)

## Video transcript

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.