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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 (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.
• Why did he include just the Discriminant part in the case of the hyper bola at ?? 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
• In why 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!