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## Precalculus (2018 edition)

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

Lesson 12: Challenging conic section problems (IIT JEE)

# 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.
• 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.
• is it correct that maths in india has very wide syllabus and reach as compared to foriegn.
• 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.
• 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.