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## Precalculus (2018 edition)

### Course: Precalculus (2018 edition) > Unit 2

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 (4 of 5)

2010 IIT JEE Paper 1 Problem 45 Circle Hyperbola Common Tangent Part 4. Created by Sal Khan.

## Want to join the conversation?

- Wait are they really expected to do all this madness under a time crunch and without a calculator? I solved this video on my own, but used a calculator... Without one this would be ridiculous.(6 votes)
- See my comment upper. There is an easier solution. Don't forget they suggest you 4 line equations and you simply have to check which line equation satisfy the tangency conditions. It takes 5 minutes. To bad Sal didn't see it. Anyway, this is a good practice but they surely don't expect this 5 videos solution at the exam.(9 votes)

- is it correct that maths in india has very wide syllabus and reach as compared to foriegn.(2 votes)
- How does ITT JEE compare with IMO?(1 vote)
- It is like the difference between a driving test (JEE) and a Formula One competition (IMO). Chalk and Cheese!

JEE Maths appears to be multiple-choice standardized stuff that most people doing mathematics at high-school can solve in a matter of minutes. The problems typically have a standard solution that the student is expected to recall and apply.

Maths in the International Mathematical Olympiad involves questions that do not have a standard solution and which competitors have an hour or more to solve. Solutions typically require some insight that the competitor will not have come across before. Few high school students would be able to solve IMO problems at all, let alone in the time available.(3 votes)

- So there is a faster way to solve this problem. If you know your polar coordinates, then you know that the circle can be given by r=8cos(Θ)*. Moreover,
**dy/dx = (sin^2(Θ)-cos^2(Θ))/(2sin(Θ)cos(Θ))* from (http://tutorial.math.lamar.edu/Classes/CalcII/PolarTangents.aspx) Then using slope-point form of a line, we know the equation of the tangent line is:**

*y = dy/dx(x-rcos(Θ))+rsin(Θ)*

We know *r=8cos(Θ)* and *dy/dx, so replacing that with our equation and simplifying, we should obtain:

y = (sin^2(Θ)-cos^2(Θ))/(2sin(Θ)cos(Θ))(x-8cos^2(Θ))+8sin(Θ)cos(Θ)*

Yes, this is kinda messy, but it works out nicely. At the end you should get:

*y = -xcos(2Θ)/sin(2Θ)+4cos(2Θ)(1+cos(2Θ)/sin(2Θ)+4sin^2(Θ)/sin(2Θ)*

*= -xcos(2Θ)/sin(2Θ)+(4cos(2Θ)+4cos^2(2Θ)+4sin^2(Θ))/sin(2Θ)*

*= -xcos(2Θ)/sin(2Θ)+4(cos(2Θ)+1)/sin(2Θ)*

Since 2Θ is present in all the functions, we can replace it with Θ. Thus:

*y = -xcos(Θ)/sin(Θ)+4(cos(Θ)+1)/sin(Θ)*

Great, that's our tangent line to the circle! If you know your formulas, you should be able to derive that very quickly.

Now the tangency constraint of a hyperbola is *c^2=9m^2-4, when c and m represents the y-intercept and slope of the tangent line, respectively. Sal Khan derived this from https://www.khanacademy.org/test-prep/iit-jee-subject/iit-jee/v/tangent-line-hyperbola-relationship-very-optional.

We just derived our tangent line to the circle, so we know our m and c! By substitution, we have:

*(4(cos(Θ)+1)/sin(Θ))^2 = 9(cos(Θ)/sin(Θ))^2 - 4

16(cos(Θ)+1)^2/sin^2(Θ) = 9cos^2(Θ)/sin^2(Θ) - 4

We can multiply both sides by sin^2(Θ) to remove the denominators:

16(cos(Θ)+1)^2 = 9cos^2(Θ) - 4sin^2(Θ)*

Multiplying the left-hand side results in:

*16cos^2(Θ)+32cos(Θ)+16 = 9cos^2(Θ) - 4sin^2(Θ)*

We can add 4sin^2(Θ) to both sides and use the fact that sin^2(Θ)+cos^2(Θ)=1:

*12cos^2(Θ)+32cos(Θ)+20 = 9cos^2(Θ)*

Taking all the stuff on the right to the left produces the quadratic:

*0 = 3cos^2(Θ)+32cos(Θ)+20

Okay, so this is a quadratic, so we can use the quadratic formula to see that cos(Θ) has two solutions, of which only one is viable:

cos(Θ)=-2/3 OR cos(Θ)=-10 (which cannot be)*

If *cos(Θ)=-2/3 then sin(Θ)=sqrt(5)/3.

Now that we know the values of cos(Θ) and sin(Θ), recall the tangent line we derived earlier:

y = -xcos(Θ)/sin(Θ)+4(cos(Θ)+1)/sin(Θ)*

Substitution for values results in our answer!

*y=2x/sqrt(5)+4/sqrt(5)* which results in

*2x-sqrt(5)y+4=0

If you know your formulas then you should be able to solve this very quickly.

Thanks for reading, I know formatting here is not that great.(2 votes) - why would we assume (23m^2)^2 as 529m^4?(0 votes)
- It's all about rearranging the numbers to become 23*23*m^2*m^2=23^2*m^2=529m^2*m^2=529m^4(3 votes)

## Video transcript

We're in the home stretch. We figured out a constraint
on b in terms of m for the line that's
tangent to both the circle and the hyperbola. And I want to make one
clarification here. Here in the last video,
I took the principal root when I took the square
root, and that's because I want the
y-intercept to be positive. Remember, when we
look at this drawing that I did in the
first video over here, this y-intercept needs
to be positive in order for it to have a
positive slope and be tangent to both the circle
and the hyperbola in this way. So down here, you
definitely want to take the principal root. Now with that said, this is the
constraint from the hyperbola, this is the constraint
from the circle. Let's set them equal to
each other and solve for m, and then we'll have
our slope of the line. So let's set them
equal to each other. So we have the square
root of 9m squared minus 4 is going to be equal to
negative 4m plus 4 times the square root of
m squared plus 1. So the first thing
we can do, let's square both sides of this. So we get 9m squared minus 4
is equal to this term over here squared. So this is 16m squared and
then plus 2 times the product of these two terms. So the product is going
to be negative 16m. So it's going to be negative
32m times the square root of m squared plus 1, and then
plus-- square this term-- 16 times m squared plus 1. So let's try to simplify
this a little bit, get maybe the radical on
one side of the equation, and see if we can
simplify this even more. So a good place to
start, so let's see. This term right
over here is going to be 16m squared plus 16. So on the right-hand side--
let me just write it like this. So we have 9m squared
minus 4 is equal to-- on the right-hand side,
we have 16 plus 16. We have 32. 32m squared plus 16, minus
32m, times the square root of m squared plus 1. Now let's see. We can subtract 9m
squared from both sides, so we'll subtract 9m
squared from both sides. And then we can add
4 to both sides. And then we are left with--
actually, let me do it the other way, because
I want to isolate this on the right-hand side. Let me do it the other way. So let's subtract 32m
squared from both sides. Well, yeah, might as
well do it this way. So subtract 32m squared
from both sides, and then subtract
16 from both sides. The left-hand
side, 9 minus 32 is going to be negative
23, because 23 plus 9, yup, is-- so this is
negative 23m squared minus 20 is equal to negative 32m
times the square root of m squared plus 1. Now we can square both
sides of this equation. It's not the cleanest
problem in the world, but hopefully, if we
haven't made any mistakes, we'll get someplace productive. Actually, let's
multiply both sides of this equation times negative
1 just to make it positive and simplify things. So that becomes positive,
positive, and positive. If you square the left-hand
side of the equation, we get 23 squared. Let me write it this way. We get 23m squared squared plus
2 times the products of this. So this would be 40--
let me just write it 2 times 20 times 23m squared,
plus 20 squared, which is 400, is going to be
equal to 32 squared, or I'll write it as 32m
squared, times m squared plus 1. Now let's see what
we can do over here. So this turns into-- this is
just a big tedious problem. So I'll just write this
is 23m squared squared plus-- so this is
going to be 40. This is 40 times 23. So 23 times 4 is 92, so it's
going to be plus 920m squared. Actually, let me expand
everything out here. Let me just go straight to
the numbers, so 23 times 23, 23 squared. 3 times 3 is 9. 3 times 2 is 6. 69, put a 0 here. 2 times 3 is 6. 2 times 2 is 4. So you get a 9. 6 plus 6 is 12. 1 plus 4 is 5. So you have 529m to the
fourth plus 40 times 23. 4 times 20 is 80. 4 times 23 is 92. So it's going to be
plus 920m squared plus 400 is equal to-- so
we're going to have 32 squared. Let me figure out
what 32 squared is. 32 times 32. It seems like if you're
taking this exam, you might as well have your
multiplication tables memorized to about 50. And I think they
do that on purpose. So just to get a sense, I
mean, this isn't an exam where the test takers
expect everyone to get a perfect-- I think
the number one score in India is on the order of 80%,
which is pretty darn amazing. To be able to do roughly the
math section in about an hour, to 80% right is pretty amazing. But I think to make the
threshold of getting into these highly
selective colleges, I think you have to get
around 40% or 50% correct. Forgive me if I got
that number wrong, but I think it's on that order. So they need to do
that because you have 200,000 or 300,000
people in India taking it, so you need to
make sure that you have a good spread of people. So this is what they do
to spread people out. So we have 2 times 2 is 4. 2 times 3 is 6. And then put a 0 here. 3 times 2 is 6. 3 times 3 is 9. And for example, if I was
preparing for the exam, I might even go ahead
and-- you know me, I hate to memorize
things, but if you had to do this type of problem
on the order of five minutes, you might want to memorize these
type of things ahead of time. The y-intercept for a
line that intercepts a hyperbola, y-intercept for a
line that intercepts a circle, who knows? But anyway. But in general, I don't think
it's good to memorize in life. Because when you're
actually doing math problems in your life, what
matters more is you understand the
underlying meaning. You normally have plenty of
time to do it in your real life. These exams are kind of an
artificial circumstance. But anyway, 4 plus 0 is 4. 6 plus 6 is 12. So it's 1,024, is
equal to 1,024. That's the 32m squared
times m squared plus 1. Or another way, this is 1,024m
to the fourth plus 1,024, squared. And now we are in
the home stretch. So let's subtract 529m to
the fourth from both sides. So minus 529m to the fourth,
this is extremely tedious. Minus 529m to the fourth. Frankly, at this
point in the problem, you're better off if you
just want to do it for speed, just trying out the
choices they gave you and figuring out which
m's and b's satisfy that. But anyway, let's
just move forward-- or which m's satisfy it. That would probably be faster. But let's just
solve it properly. So minus 529m to
the fourth plus-- and then we also want to
subtract 920m squared. So let's also
subtract 920m squared. And so on this side,
we're just left-- let's also subtract
a 400 just to make this a proper quadratic. And also we're going
to put a 400 here. So the whole left
side simplifies to 0 is equal to-- 1,024 minus 529. Let's see. 1,024 minus 524
would have been 500. And so it's going to
be 5 less than that. So it's going to be
495m to the fourth. Did I do that right? 1,024 minus 524
would be 500, but I'm subtracting 5 more than that. So that is going to be 495. And then 1,024 minus
920, it'll be 80 plus 24. So it's going to be plus 104m
squared minus 400 is equal to 0 So we have just a straight-up
quadratic equation over here. And you might not recognize
it, but this is the same thing. M to the fourth is the same
thing as m squared squared. So let's just solve it. So we'll get m squared. We're not solving for x anymore. We're going to get m squared
using the quadratic formula. You could substitute x
is equal to m squared, and this will just become
a natural quadratic then. m squared is equal to negative
b, negative 104, plus or minus the square root-- more fun math
for us without a calculator-- plus or minus the square
root of 104 squared minus 4 times a, which is 495,
times c, which is negative 400. So that'll make this a positive. And you have a 400
right over here, and then all of that
over 2 times 495. 2 times 495 is what? Well, it's going to
be 10 less than 1,000, so it's going to be 990. So let's try to see if
we can evaluate this. Actually, I'll stop
this problem here. I'm already cross the
10-minute threshold. In the next video,
we'll just grind through this
mathematics-- It's really just arithmetic at this
point-- and figure out what m is going to be equal to.