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## 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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# Common tangent of circle & hyperbola (2 of 5)

2010 IIT JEE Paper 1 Problem 45 Circle Hyperbola Common Tangent Part 2. Created by Sal Khan.

## Want to join the conversation?

- I'm watching this video in the Algebra: Conic sections topic. What does IIT JEE mean?(5 votes)
- It is the toughest exam during under graduate level in the world. (Approved)(1 vote)

- 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

equations:

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)
- No. These are IIT JEE problems. These are not problems made by Khan Academy.(3 votes)

- how do i get 90% marks in jee?(1 vote)
- Iam preparing for jee. Is it help me in SAT?(1 vote)

## Video transcript

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.