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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 (3 of 5)

2010 IIT JEE Paper 1 Problem 45 Circle Hyperbola Common Tangent Part 3. Created by Sal Khan.

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- 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 did he include just the Discriminant part in the case of the hyper bola at3:43?? Where as in case of the circle he considered the whole formula....??(1 vote)
- he used the discriminant part in the previous video too

just re-watch the 2nd part and you will understand that he uses the whole formula in the later part of the video when he is solving for 'b' in terms of 'm', from the equation he got by equating the discriminant(2 votes)

- In3:52why doesn't he multiply everything with -1 first in order to have only one negative? Is there a reason not to do that? Thanks in advance!(0 votes)
- It doesn't really matter. It's not like he cancelled out all of the negatives, so it's a waste of time.(2 votes)

- how much time do they give us to answer it?(0 votes)
- Average time given per question is 3 minutes.(0 votes)

- At4:20why was Sal able to square -18mbx without squaring anything else?(0 votes)

## Video transcript

In this video we're going
to do with the hyperbola the exact same thing
we did with the circle. We're going to find
constraints on the y-intercept for the tangent
line in terms of m. But this time we're going
to use the hyperbola. And then we can set them
equal to each other, and solve for the m. So let's remind ourselves what
the equation of the hyperbola is. They give it to us
right over there. It's x squared over 9 minus y
squared over 4 is equal to 1. Let me write this over here. So it is x squared over
9 minus y squared over 4 is equal to-- I'll
write the y squared-- minus y squared over
4 is equal to 1. But now we can
substitute the y squared with the y is equal
to mx plus b that we got from the last video. And we actually figured out
what y squared is equal to. Y squared is equal to all
of this business up here. Because it's the same line. Remember, this is the
whole point of it. We're trying to find two
constraints on that same line. So we can rewrite
this same thing. And actually one
thing I want to do is I'm going to multiply
both sides of this equation by 36, the common
multiple of 9 and 4, so that I can get rid
of these fractions. And so this is going to become
36-- let me write it this way-- so 36 divided by 9 is 4. So it's 4x squared minus--
36 divided by 4 is 9-- and I would put a y
squared here but we know that y squared is
the exact same thing as this thing over here. So y squared is the
same thing as m squared. x squared plus 2
mbx plus b squared. And then this is going to
be equal to-- remember, we multiplied both
sides of this equation by 36-- so this is
going to be equal to 36. And let me simplify this. And we're going to do
the exact same thing. We know that the m and
b have to be such that, or the line has to have a slope
and y-intercept where it'll only intersect with the
hyperbola at one point. It'll only have one solution to
this quadratic in terms of x. But let's simplify it first
before we worry about that. So this is equal to 4 x
squared minus 9 m squared x squared minus 18
mbx, minus 9 b squared. Just multiply it there. And let me subtract
this 36 from both sides, so then we have minus
36 is equal to 0. So this is a quadratic
in terms of x. But let me combine the
various degree terms. So these are the x squared
terms right over here. So this is the same thing as
4 minus 9 m squared times x squared. And then our only x term
is this right over here. That's our only x term. So this is minus 18 mbx. And then our constant
terms are right over here. So this over here is minus
9 b squared-- and let me write it this way, let
me write it as minus 9-- well I'll just write it,
minus 9 b squared minus 36. I'll just write it like that. And of course that is
going to be equal to 0. And remember, quadratic
formula, if we wanted to solve for the Xs,
we'd have the quadratic formula, but we only want to
have one solution. So the discriminant part
of the quadratic formula is going to equal 0. The b squared minus 4 ac
is going to be equal to 0. This is exactly what we
did in the last video. So let's take the
b squared minus 4ac and set that equal
to 0, and then we'll have our constraints on m and b. So b squared is-- and
remember don't get the squared in the quadratic
formula confused with the b in the y-intercept. But this term
squared is-- so this is going to be-- 18 squared
m squared b squared, right? Negative 18 squared
is just positive 18. Minus 4 times a. a is 4 minus 9 m
squared times c. So I can rewrite c as negative
9 times b squared plus 4. Did I do that right? Negative 9 times b squared
is negative 9 b squared. Negative 9 times
4 is negative 36. I want to make sure I don't
make any careless mistakes. And so this becomes--
so if we just take the negative 9
and the negative 4-- they become a positive 36. They become a positive 36. And then we can actually,
just to simplify things so we don't have to do
too much fancy math, this 18 squared--
let's remember-- 18 squared is the same
thing as-- well 18 squared is going to be divisible--
well I won't worry too much about that just yet. I just want to make
sure that-- Actually let me write it this way. So 18 squared is 2
times 9 times 2 times 9. Or another way to think about
it is 4 times 9 times 9. That's the same
thing as 18 squared. 4 times 9 times 9. Now we can divide both of these
terms-- and this whole thing, remember we want this whole
thing to be equal to 0. The discriminant has
to be equal to 0. So we can divide both
sides of this equation by 36, which is the same thing
as dividing by 4 times 9. So this term right over here,
we could get rid of one 4 and one 4. And we're going to get a 9 m
squared b squared over here. And then we divided by
36, so these all go away. So it's going to be
9 m squared b squared plus this thing
times this thing. So let's see what that is. So we have a 4
times a b squared. So we have plus 4 b squared. Let me do this in
a different color. I'll do it in blue. So plus 4 b squared. And then you have 4 times 4. So plus 16. And you have negative 9 m
squared times b squared. So it's negative 9
m squared b squared. And then you have negative
9 m squared times 4. So negative 36 m squared. And that's going
to be equal to 0. Lucky for us, that
and that cancel out. And then we are
left with something. And actually, what
we're left with, everything is divisible by 4. So let's divide what
we're left with by 4. So then we're left with b
squared-- that's that term. And then minus 9 m squared. That's that term over there. Just divided it by 4. And then plus 4 is equal to 0. And once again, we could
use-- well actually here, we don't have to do anything. We don't have to use
quadratic formula. We could just solve for b. We could subtract
this from both sides. And so we'll get b squared
is equal to the square root of 9 m squared minus 4. Sorry-- let me
just write this-- I don't want to skip steps here. b squared is equal to
9 m squared minus 4. They're painting the
office right now, so maybe it's making me a
little bit-- making my brain not work properly. Then b is equal to the square
root of 9 m squared minus 4. Did I do that right? Let's see. I got the 4. Yeah, it looks right. And so we're left
with a situation where the b-- if we're saying
that if the line is tangent to the
hyperbola it to be this, and if the line is
tangent to the circle b has to be equal to this,
this business over here. So let me copy it and
then let me paste it. Let me paste it. Just like that. And so we now have two
equations with two unknowns. We can set these equal to
each other and solve for m. And this will give us the m or
the slope of that tangent line. And then we can go
ahead and solve for b. I'll do that in the next video.