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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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# Common tangent of circle & hyperbola (1 of 5)

2010 IIT JEE Paper 1 Problem 45: Find the equation of a tangent common a given circle and a given hyperbola. Created by Sal Khan.

## Want to join the conversation?

- At4:05, what does he means when he said the hyperbola is going to have a higher slope than the asymptote? I thought the slope of the horizontal hyperbola is always lower than the asymptote because it can never touch it.(6 votes)
- If the slope of the hyperbola line was lower than the slope of the asymptote, it would pull away from the asymptote. Alternatively, if the slope was the same between the two, they would be parallel lines and so the hyperbola line would not approach the asymptote.

The slope of the hyperbola line has to be higher than the slope of the asymptote. As the hyperbola gets closer and closer to the asymptote then the slope of the hyperbola get closer and closer to the slope of the asymptote. This ensures that the lines can never touch.(14 votes)

- Hey guys! Are you aware of any faster ways to solve this problem? I am afraid that there is simply not enough time to go through this rigid solution at the actual exam.

I will post an alternative approach below. Please feel free to comment whether it seems faster and provide any other creative ways! :)(10 votes)- You can note that since there just one intersection between the circle

x^2+y^2-8x=0 and the straight line y=mx+c, if you were to make these two equations equal the discrimination of the resulting quadratic should be zero. Similarly, the quadratic upon solving for the intersection between the hyperbola and the straight line must also have a discriminatory of zero. This gives two simultaneous equations in m and c which can be checked against the options or solved if you want to do things the proper way.(1 vote)

- Is there any shorter method also? Like calculus or anything else? Please explain?(3 votes)
- Elimination, as this is a multiple choice question would work.(4 votes)

- Is it accurate to say that the tangent can't have a lower slope than the positive asymptote else it would intersect the hyperbole twice and no longer be a tangent to the hyperbole?(4 votes)
- At3:33he says that it can't be top right and bottom left circumference of the circle, but I don't understand why is that? And why is top left and bottom right can be the part we want?(1 vote)
- It can't be top right or bottom left because the tangent needs to have a positive slope, that is to look like a forward slash /, not a backslash \. The bottom right part is impossible to have a common tangent with the hyperbola, so the only part of the circle possible to draw the tangent on is the top left.(6 votes)

- Hi!

I tried to solve this problem using Calculus, but I probably do not even know what I am doing since all the Calculus I know is from watching videos. I haven't taken a Calculus course yet (I am a Freshmen in high school), and I only know derivatives.

This is my approach: The question asks for a common tangent for the circle and hyperbola. That means that the same line has to intersect the circle and hyperbola. As a result, the line will have the same slope.

Using this reasoning, I took the derivative of the circle and the derivative of the hyperbola equal to get a common slope. I simplified and got a quadratic solution, whose 2 solutions are a point on a hyperbola and a point on the circle. I then tried to use the points to determine an equation.

The solution I got for the points had crazy decimals, so it would be cool if someone could tell me what is wrong with my reasoning.

Thanks!(2 votes)- If I understand your intuition and steps correctly, I think the mistake was made when you took the derivative of the circle and set it equal to the derivative of the hyperbola. You are correct in saying that both tangent lines will have the same slope, but that does not necessarily mean that they are the
*same*line, as there could be different lines with the same slope. For example, take sin(x): there are many points where the derivative of sin(x) is 1. Setting the derivative of sin(x) equal to 1 would give you x = 0, 2pi, and so on, which in the context of the problem might not be what you want to find.

I, too, only have experience with Calculus from what I have learned in my own exploration of math; therefore, I could be wrong, but I think that is the mistake in your approach.(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.(3 votes) - Why did they include the part about the intersection at points A and B? Does that mean anything to the problem?(2 votes)
- Yes and no. The reason it DOES mean something is basically because it aids in drawing the sketch, since we're not given one. But other than that, you don't use it in the problem.(3 votes)

- Find Equations of tangents through the given point:

i) x^2 + y^2 =25 through (7,-1)

ii) y^2=12x through (1,4)(1 vote) - there's is something i think is weird that i got.... First i considered one tangent point to both he circle an hyperbola, and i got c (in y intercept of the tangent line) as an imaginary number 2i(squareroot of 15)/3 . is that even possible?(1 vote)
- How did you get that? Can you be more specific?(3 votes)

## Video transcript

The circle x squared plus y
squared minus 8x equals 0, and the hyperbola x squared
over 9 minus y squared over 4 equal 1
intersect at the points A and B. Equation of a common
tangent with positive slope to the circle as well so
the hyperbola is-- so let's just visualize what
they're asking first. And this is going to
take us multiple videos, I have a feeling. But let's just
visualize it just so we can get our head
around the problem. So this circle, let me complete
the square in terms of x. So this circle, as they wrote
it, is x squared minus 8x. And then we have
a plus y squared-- I left some space here
so we can complete the square-- is equal to 0. And then let me add half of this
8 term squared to both sides. So half of negative
8 is negative 4, and negative 4 squared is 16. So add 16 to both sides. And that allowed me to turn the
x-term into a perfect square. This is the same thing
as x minus 4 squared. And then we have plus y squared. Plus y squared is equal to 16. So this is a circle. This right here is a circle
with center at x equals 4. At x equals 4, y is equal to 0. And it has a radius
of 4 as well. So let me graph
this circle here. So let me draw the
horizontal axis, my x-axis. Let me draw the y-axis. That is my y-axis over here. And let me draw its
center, so 1, 2, 3, 4. 4 comma 0. That's it's center, and
it has a radius of 4. So it's going to
come out, and it's going to look something--
I could draw a better circle than that. It's going to look something
like-- that's the top half, and then the bottom
half is going to look something like that. So that's our circle. Now let's think
about the hyperbola. So if we just look at it, the
x squared term is positive, so it's going to
be a hyperbola that opens to the right and the left. We do this a bunch in
the conic sections videos if you want to review of that. And we could just figure out
where it intersects the x-axis. So then when y is equal to
0, we have x squared over 9 must be equal to 1. Or x would be plus or minus 3. So the hyperbola is
going to look like this. So this is at plus 3 comma 0. The hyperbola will
open up like that. And then at 1, 2, 3,
negative 3 comma 0, the hyperbola is going
to open up to the left. And so in the problem when they
describe the points A and B, they're probably talking about
that point A and that point B. Now, let's think about what
this question is asking us. Equation of a common tangent
with positive slope-- so it has to have a positive
slope-- to the circle as well as to the
hyperbola-- a common tangent. So let's just think
about this a little bit. So it's going to have
a positive slope, so it won't be
tangent to the circle anywhere where the circle
has a negative slope. So it can't be
tangent over here. It can't be tangent over there. And then we could
say, well, if it was tangent to the circle
over here, what would happen? Well , it wouldn't be able to
be tangent to the hyperbola. So it has to be
tangent to the circle someplace in this blue
region right over here. And then how can it be
tangent to the hyperbola? It might be tempting
to say that it would be tangent to the hyperbola
in this way somehow, but what you need to
realize is the hyperbola is asymptoting
towards some line. And we could figure
out what that line is. Its asymptoting
towards some line. So let me draw that line. The hyperbola is always
going to have a higher slope than that line, a
very slightly higher slope. It's slowly
approaching that line. So if you go out
here, the hyperbola is going to have a higher
slope than the asymptote line. And so if you had
to be tangent to it, you would have to
have a higher slope. And anything that's
coming from this part of the circle towards anything
out here on the hyperbola is going to have to have a lower
slope than the tangent line, right? Because it's going
to have to catch up. The tangent line
is going to have to catch up to
whatever-- let me draw it again-- to whatever
we draw over here. I want to make this clear. The hyperbola, as
you go out here, actually this whole period, this
whole part of the hyperbola, is going to have a
higher slope than what it is asymptoting towards. That's what allows it to get
closer and closer to that line. So any tangent is going
to have to have a higher slope out here. Anything tangent
to the hyperbola is going to have
to have a higher slope than this actual line. It's going to have to have
a slightly higher slope. So if we take something
out here and we try to draw a tangent from this
part of the circle out there, it wouldn't work. Because this tangent,
by definition, in order to meet
the hyperbola, is going to have to have a lower
slope than this asymptote. So this can't be tangent to
that part of the hyperbola. So what else can we do? Well, the only other
part of the hyperbola that we might be able
to work something out is this part of the
hyperbola right over here. So if we find a line that's
tangent maybe there and there on the hyperbola, then we might
have found our common tangent with positive slope. So let me draw that. So our common tangent
with positive slope, I'll do that in pink. Our common tangent with positive
slope could look like that. So now that we have
the visualization down, in the next video
let's try to figure out what that line might look like,
especially when we constrain it to having to be
tangent to the circle and having to be tangent
to this hyperbola.