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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 (5 of 5)

2010 IIT JEE Paper 1 Problem 45 Circle Hyperbola Common Tangent Part 5. Created by Sal Khan.

## Want to join the conversation?

- There must be a very shorter way to solve the question unlike this solution. Can you make a tutorial on shorter ways to solve this problem ? Because we definitely are given
**3 hours**to solve the whole question paper.(9 votes)- You can find the tangent slope vs. y-intercept equation for both the hyperbola and circle pretty easily, a minute each maybe, then all you need to do is plug in the slope and y-intercepts of the different answer choices until they're equal. Sal made this way harder(2 votes)

- Wow... what is the time limit in the JEE exam to solve this problem?(3 votes)
- 2.5 min MAXIMUM!(15 votes)

- Is there a faster way to do this problem using calculus?(1 vote)
- Joseph,

There is a faster way, but it does not require any calculus knowledge.

In general, when you solve problems under time constraints and you are provided with answer choices, one potential technique is to check whether each answer choice is a solution to the problem at hand. According to this approach, you should not try to find general solution to the problem and check which answer choice corresponds to your findings (what Sal did for this question), but rather try to go the other way around: substitute each answer choice to your equations and check which answers satisfy the constraints of the problem. This technique may often help you to find correct answers faster.

Let us apply this approach to our issue. First, we need to ensure that we fully understand the situation and the objective. In this case we do so by plotting the circle and the hyperbola as Sal did at the beginning of the corresponding video: https://www.khanacademy.org/math/algebra2/conics_precalc/jee_hyperbolas/v/iit-jee-circle-hyperbola-common-tangent-part-1.

It is clear that we are looking for a line that is tangent to**both**the circle and the hyperbola. This means that the line should have**exactly one common point with both**the circle and the hyperbola.

After understanding the question, let us look at our answer choices. We need to essentially find an equation of a tangent, which is a line. Therefore, we would like to rewrite each of the answer choices in a standard form, so that we could quickly analyze them:`(A) y = (2⁄√5)⋅x − 20⁄√5`

`(B) y = (2⁄√5)⋅x + 4⁄√5`

`(C) y = (3⁄4)⋅x + 2`

`(D) y = (4⁄3)⋅x + 4⁄3`

By the logic explained at around9:15in https://www.khanacademy.org/math/algebra2/conics_precalc/jee_hyperbolas/v/iit-jee-circle-hyperbola-common-tangent-part-2, the y-intercept of our tangent should be positive. If we look at our answer choices with this in mind, we can immediately recognize that choice (A) is not correct. It is the only answer that has a negative y-intercept of −20⁄√5. As our y-intercept should be positive, we can automatically**eliminate choice A**.

As you can see, by simply looking at the problem and slightly modifying the answer choices, we have already been able to eliminate one of the answers, raising our probability of guessing correctly from 25.00% to ≈ 33.33%.

The next step should be to check whether our 3 potential candidates for a tangent**satisfy the first condition**: they should have**exactly one common point**with our circle. A tangent is a line and so it has an equation of**y = mx + b**. Instead of finding**m**and**b**that satisfy our constraints (Sal has already done this for us), we will take pairs of**m**and**b**from our answer choices and check whether these pairs satisfy our constraints.

As an example, for choice (B): y = (2⁄√5)⋅x + 4⁄√5,**m**= 2⁄√5 and**b**= 4⁄√5.

To check whether the answer choices satisfy the first condition of having**exactly one common point**with our circle, we may simply substitute our**y**from the equation of a tangent**y = mx + b**to the equation of the circle (Sal explains it at around1:25in https://www.khanacademy.org/math/algebra2/conics_precalc/jee_hyperbolas/v/iit-jee-circle-hyperbola-common-tangent-part-2):`y = mx + b`

– equation of the tangent`x² + y² − 8x = 0`

– equation of the circle`x² + (mx + b)² − 8x = 0`

– equation for x where the tangent intersects with the circle

In order for our tangent and our circle to have**exactly one common point**, this new equation**x² + (mx + b)² − 8x = 0**should have only one solution. This is a quadratic equation, hence what we really need to do to check if an answer choice satisfies the condition is to plug corresponding**m**and**b**from this answer choice to the equation above and to test whether the discriminant of the resulting equation is 0.

We can work through the choice (B) together, so that you can see what I am talking about. The first step is to substitute**m**and**b**from choice (B) into the equation of our first condition:`x² + (mx + b)² − 8x = 0`

`x² + ((2⁄√5)⋅x + 4⁄√5)² − 8x = 0`

Simplifying:`x² + ((2⁄√5)⋅x)² + 2⋅(2⁄√5)⋅x⋅(4⁄√5) + (4⁄√5)² − 8x = 0`

`x² + (4⁄5)⋅x² + (16⁄5)⋅x + 16⁄5 − 8x = 0`

Multiplying both sides of the equation by 5:`5x² + 4x² + 16x + 16 − 40x = 0`

`9x² − 24x + 16 = 0`

Checking the discriminant:`D = 24² − 4⋅9⋅16 = (8⋅3)² − 4⋅3⋅3⋅8⋅2 = 8⋅3⋅8⋅3 − 8⋅3⋅8⋅3 = 0`

The discriminant of the simplified equation of our first condition is indeed 0, which implies that there is only one**x**where our tangent and our circle intersect. Therefore, choice (B) does satisfy the first condition and cannot be eliminated at this stage.

Note that we did not need to solve for**x**in the equation above, we just needed to ensure that there is only one**x**that satisfies our equation. To do this we just check whether the discriminant is equal to 0.

If you substitute two other relevant answer choices (C) and (D) into the equation of our first condition

(x² + (mx + b)² − 8x = 0), you will see that for both answer choices the discriminants are also equal to zero. Therefore, all three answers that are left after we eliminated choice (A) satisfy the “exactly one common point with the circle” condition, and neither can be eliminated at this stage. This is unfortunate, but we must not despair!:) We just need to remember that there is another condition to test.

As you remember, our tangent line should not only have**exactly one common point with the circle**, but should also have**exactly one common point with our hyperbola**, which is going to be our**second condition**.

We can simplify the equation of the hyperbola:`x²⁄9 − y²⁄4 = 1`

Multiplying both sides of the equation by 36:`4x² − 9y² = 36`

Now, let us substitute our**y**from the equation of a tangent**y = mx + b**to the equation of the hyperbola:`y = mx + b`

– equation of the tangent`4x² − 9y² = 36`

– equation of the hyperbola`4x² − 9⋅(mx + b)² = 36`

– equation for x where the tangent intersects with the hyperbola

At this stage, we just need to substitute three answer choices that are left into this equation and find for which of them the equation has exactly one root. A single root means that our hyperbola and our tangent have**exactly one common point**, which satisfies our second condition. Let us work through the process for choice (B) first:`4x² − 9⋅(mx + b)² = 36`

`4x² − 9⋅((2⁄√5)⋅x + 4⁄√5)² = 36`

`4x² − 9⋅[((2⁄√5)⋅x)² + 2⋅(2⁄√5)⋅x⋅(4⁄√5) + (4⁄√5)²] = 36`

`4x² − 9⋅[(4⁄5)⋅x² + (16⁄5)⋅x + 16⁄5] = 36`

Multiplying both sides of the equation by 5:`20x² − 9⋅5⋅[(4⁄5)⋅x² + (16⁄5)⋅x + 16⁄5] = 180`

`20x² − 9⋅[4x² + 16x + 16] = 180`

`20x² − 36x² − 144x − 144 = 180`

Simplifying and subtracting 180 from both sides of the equation:`−16x² − 144x − 324 = 0`

Multiplying both sides of the equation by −1:`16x² + 144x + 324 = 0`

Looking at the discriminant:`D = 144² − 4⋅16⋅324 = (12⋅12)² − 4⋅4⋅4⋅9⋅36 = (12⋅12)² − 12⋅12⋅4⋅3⋅12 = (12⋅12)² − (12⋅12)² = 0`

The discriminant for choice (B) is indeed equal to 0 and, therefore, our hyperbola and the tangent represented in choice (B) intersect at exactly one point, which is what we are looking for.

We showed that the tangent described in choice (B) has**exactly one point of intersection with both our circle and our hyperbola**and has a positive slope. It satisfies all the conditions of our problem and, therefore,**choice (B) is the right answer**!

It is also worth working together through the second condition using the answer (C):`4x² − 9⋅(mx + b)² = 36`

`4x² − 9⋅((3⁄4)⋅x+2)² = 36`

`4x² − 9⋅[((3⁄4)⋅x)² + 2⋅(3⁄4)⋅x⋅2 + 2²] = 36`

`4x² − 9⋅[(9⁄16)⋅x² + 3x + 4] = 36`

`4x² − (81⁄16)⋅x² − 27x − 36 = 36`

`(64⁄16)⋅x² − (81⁄16)⋅x² − 27x − 72 = 0`

`−(17⁄16)⋅x² − 27x − 72 = 0`

`(17⁄16)⋅x² + 27x + 72 = 0`

Looking at the discriminant:`D = 27² − 4⋅(17⁄16)⋅72 = (3⋅9)² − 4⋅(17⁄16)⋅4⋅18 = (3⋅9)² − 17⋅18 ≠ 0`

The discriminant in this case is clearly not equal to zero as (3⋅9)² does not contain 17 as a factor, but 17⋅18 does. In fact, the discriminant is positive (if you are not certain about this, simply calculate it, you will get 423), which means that the line described in choice (C) has**two points of intersection with our hyperbola**. If a line has two points of intersection with a hyperbola it cannot be its tangent by definition. Choice (C) does not satisfy the second condition of our problem and, therefore,**choice (C) is NOT the right answer**.

You can go through the same process to show that**choice (D) is NOT the right answer**as well, because when we plug the equation of the line from choice (D) into 4x² − 9⋅(mx + b)² = 36, we eventually get to a quadratic equation with the negative discriminant.

Overall, by simply plugging answer choices into the conditions of this problem and doing relatively easy algebra, we managed to avoid all the hairy calculations that we encountered together with Sal when trying to get to a general solution. This approach may save you some precious time during the actual exam and help to avoid unnecessary mistakes.

Please do not hesitate to ask any questions about this problem or to provide any comments.

Have fun!:)(10 votes)

- A better way to solve this question would be by brute force method. If you check the options, only option (b) is actually tangent to the circle. Option (a) is directly eliminated as it has a negative y-intercept - which isn't possible. So the answer is clearly option (b). But of course, knowing the actual way to solve is also important. :)(8 votes)
- Just saying, but at roughly2:00in the video, those 10s were divisible by 2, and since there where two 10s and two 2s, Sal could've factored the whole thing by another 4 on both sides before adding the stuff in the square root and continuing factoring... Just saying...(3 votes)
- This question was crazy.(2 votes)
- When you see a test problem like this that is this horribly nasty, there is usually a quicker solution using underlying principles. Other comments on this video have mentioned better ways to solve this faster. :)(2 votes)

- Sounds like he was out of breath.(2 votes)
- I think there's a better way to calculate the square root, like 50176, the last three numbers are 176, we can use 176 to minus the first two numbers 50, 176-50=126, and 126 is divisible by 7, so we can divide by 7 directly, it's easier and faster to calculate.(2 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,**, so replacing that with our equation and simplifying, we should obtain:

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

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**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)***then

If *cos(Θ)=-2/3**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 we couldn't use the equation c^2+b^2=a^2m^2 once for the tangent and the hyper and once for the tangent and the circle? then we got 2 equations of two unknowns (m and c of the tangent).

Reference: a , b as the hyper radii and m , c are the tangent line slope and y-intercept "respectively".

I MEAN: you made almost the same thing (you did use the basics of c^2+b^2=a^2m^2) of the hyper and the tangent to find that c^2=9m^2-4. however, this gave us an equation of m and c and as we need another equation of m and c why we didn't use c^2+b^2=a^2m^2 for the circle and the tangent? we can write the equation of the circle in the form of the equation of the hyper as ((a^2)/16)+((b^2)/16)=1 and then plug these values in c^2+b^2=a^2m^2 and find the 2nd equation we need between m and c. two equations and two unknown, and that's it. is this wrong? if so, why?(1 vote)

## Video transcript

All right. Where we left off we really
were in the home stretch. And now we just have
this messy arithmetic that we have to solve to
essentially find our slope. And then we'll be able to
figure out the y-intercept. And since we're actually going
to take the square root of this eventually, let's just
factor each of these terms and see if this
hopefully will simplify into something reasonable. So let's just factor 104. I think it'll be
useful just have each of these guys factored. So 104 is the same thing as
4 times 4 times 26, right? 4 times 25 is 100. 4 times 26, which is the same
thing as 2 times 2 times 13. So that's 104. So 104 squared. Let's just worry about
the radical right now. So this is going
to be this squared. So it's going to be 4 times 4
times 2 times 2 times 13 times 13. That's this right over here. I squared it, so I put
each of its factors twice. And then you're
going to have plus 4 times-- let's factor 495. 495 is divisible by 5. Let's see. 5 goes into 500 100 times. This is five less than that,
so it's going to be 99. 99 is 9 times 11. So the 495 can be factored
as 5 times 9 times 11. And then 400. 400's pretty straightforward. 400 is going to be times
4 times 10 times 10. And so we can factor out--
so this expression right here under the radical sign, there
is a common factor of 4 times 4 here. We have a 4 times
4, and there doesn't look to be any
other common factor. So this is going to
be the square root of 16 times the
square root of-- it looks like we're
actually going to have to do some serious
multiplication here. The square root of 4 times 169. So let's just multiply that out. This is really a
painful problem. Let's see. 13 times 13 is 169. We're going to have 169,
that's the 13 times 13 times 4. So 4 times 9 is 36. 6 times 4 is 24, plus 3 is 27. 1 times 4 is 4, plus 2 is 6. So all of this right over
here is going to be 676. And then we're going
to have, let's see, 5 times not-- well we know
what this is right over here. This number right over
here is going to be 495. And we're going to
multiply that times 100. So this is going to be plus
495 and just add two 0's. , One two. So it becomes 49,500. So this is going to become--
the square root of 16 is 4, times the square root--
this will be interesting. We're going to have to do
some serious factoring here-- of 49,500 plus 676 is
going to be 50,176. If this was just a 500 here,
then we would get to 50,000, and we have another 176. So times the square root
of this crazy thing. And so let's see if
we can factor this. This is really, really
a nasty problem. So 50,000 looks like
it's divisible by 4. So let's see. 4 goes into 50,176. We're getting some good
practice here on our arithmetic. 4 goes into 50-- maybe
there's a faster way to do this-- 12 times. 12 times 4 is 48. Subtract, you get 21. 4 goes into 21 five times. You get a 20. Subtract, you get a 17. 4 goes into 17 four times. 4 times 4 is 16. Subtract, bring down,
then you get a 16. So it goes 12,544 times. And once again this looks
like it's divisible by 4. It definitely is divisible by 4. And let's see. This is going to
be-- I don't want to make any careless mistakes,
so let me just divide it again. I'm tempted to do some
of this in my head, but we've gotten so far I
don't want to make mistakes. So 4 goes into 12 three times. 3 times 4 is 12. 4 goes into 5 one time. 1 times 4 is 4. You have 14. 4 goes into 14 three times. 3 times 4 is 12. You get a 2, 24. So 3136. Once again, this looks like
it's divisible by 4 again. I should have just guessed
that this whole thing was going to be divisible by
16 or something. So 4 goes into-- don't want to
make a careless mistake-- 3136. 4 goes into 31 seven times. You get a 28. Subtract, you get a 3. 33 goes into it eight times. You get a 32. Subtract, you get a 16. You go exactly four
times, no remainder. So 784. Once again, this looks
like it is divisible by 4. It's getting at
least simpler now. 4 goes into 784. 4 goes into 7 one time. 4. You get a 38. It goes here nine times. 9 times 4 is 36. You subtract, you get a 24. 196 times. And really, if you are
going to take the JEE, I do recommend being able to
do this mental arithmetic much faster because I'm clearly
not capable of doing it fast enough. So you have 196. And 196, this is divisible
by 16, I believe. Let me see. 16 goes into 196. It goes into it--
actually no it's not. So let's just do it by 4. This is 17, I believe. Let me look. My brain is fried. 196 four times. 16, 36. 4 goes into 36 nine times,
so this is 4 times 49. All right. So if we're taking the square
root of the square root of all of this business, this
is 4 times 49 over here. So this is the square root
of 4 to the fifth power. So let's just do it this way. We could just take the square
root of each of these factors. So this in blue
is going to become 4 times the square root of this. We could just take the square
root of each of its factors. So it's going to be
1, 2, 3, 4, 5 2's. So times 2 to the fifth,
which is 32, times the square root of 49, which is 7. So this term right
over here simplified to-- this whole video
has been just arithmetic. My brain really is
fried-- so 32 times 4. We have an 8. 4 times 3 is 4 times 3 is 12. And then we want to
multiply that times 7. I think you can see my
handwriting degrade. 7 times 8 is 56. 7 times 2 is 14, plus 5 is a 19. Bring up the one. 7 times 1 is 7, so it's 896. So all that stuff
in blue was 896. So let's see. All of this stuff
over here was 896. Now we have to decide
to do plus or minus. Remember, we want
a positive slope. So we want to add the 896. So we're getting m squared is
equal to negative 104 plus 896. So if we subtract 104 from
this business over here, we're going to get 792 over 990. This is what m
squared is equal to. These are both
divisible by nine. So let's divide
both of them by 9. Let me clear all of
this stuff down here. It's Just a big
arithmetic video, but hopefully we'll
get the right answer. It would be painful if I made
a careless mistake earlier on and all of this is in vain
and I have to redo the video. So let's hope that
isn't the case. So let's divide
both of these by 9. 9 goes Into 792. It goes int 79 8 times. 8 times 9 is 72. And then you get 72. So it goes 88 times. You divide that by 9 you get 88. If you divide this
by 9, you get 110. Now these are both
divisible by 11. So if you divide by
11 you get 8 over 10. 8 over 10 is the same
thing as 4 over 5. Unbelievable. It all simplified down to 4/5. So the slope of our line
m squared is equal to 4/5. Or taking the square
root of both sides, m is equal to 2 over
the square root of 5. So that's the slope of our line. And now we just have
to go back to one of these equations, whichever
one we think is simpler, and find the y-intercept. This is the simpler one. So m is equal to 2 over
the square root of 5. b is going to be equal
to the square root of 9 times m squared. Remember m squared is 4/5. 4/5 minus 4. So what is this
going to be equal to? Now let me just clear all of
this business out here since we don't-- well I'll
just leave it there. Maybe it'll be
useful in some way if I made a careless mistake. Hopefully I haven't. So this is going to be
equal to 9 times 4 is 36, so it's equal to 36
over 5 minus 20 over 5. So this is equal to the
square root of 16 over 5. I'm running out of space. So b is equal to--
36 minus 20 is 16. So square root of 16
over square root of 5. Or 4 over the square root of 5. And we are done. We now know the slope of that
tangent line, the thing that had a positive
slope, its equation is going to be y is equal to
2 over the square root of 5x plus 4 over the
square root of 5. And if we want to simplify it,
because if I remember properly that problem didn't
have it in-- they put everything on
the right hand side. Let's multiply everything
times square root of 5. So you get square root of
5y is equal to 2x plus 4. And then we could subtract
these from both sides and we get negative 2x plus
the square root of 5y minus 4 is equal to zero. So do they have that
as one of the choices? So it looks like
they do if we just multiply everything
times negative 1. If we multiply everything
times negative 1 we get 2x minus the square root
of 5y plus 4 is equal to zero. And we are done. This was probably the
most painful problem I've done in my life. 2x minus the square
root of 5y plus 4. The answer is B.