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## Algebra (all content)

### Course: Algebra (all content)>Unit 17

Lesson 12: Challenging conic section problems (IIT JEE)

# 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?

• At , 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. • 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.
• 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! :) • 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? • 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? • At he 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) • 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! • 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.
• 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 they include the part about the intersection at points A and B? Does that mean anything to the problem?   