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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.

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  • mr pants teal style avatar for user Wrath Of Academy
    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.
    (7 votes)
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    • leafers ultimate style avatar for user Clément Amzallag
      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.
      (9 votes)
  • leaf orange style avatar for user Kingsman Agent
    is it correct that maths in india has very wide syllabus and reach as compared to foriegn.
    (2 votes)
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  • male robot hal style avatar for user King Henclucky
    How does ITT JEE compare with IMO?
    (1 vote)
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    • starky ultimate style avatar for user Sugary Toast
      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.
      (3 votes)
  • male robot hal style avatar for user Allen Lin
    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

    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)
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  • leaf blue style avatar for user Eri Tagawa
    why would we assume (23m^2)^2 as 529m^4?
    (0 votes)
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Video transcript

We're in the home stretch. We figured out a constraint on b in terms of m for the line that's tangent to both the circle and the hyperbola. And I want to make one clarification here. Here in the last video, I took the principal root when I took the square root, and that's because I want the y-intercept to be positive. Remember, when we look at this drawing that I did in the first video over here, this y-intercept needs to be positive in order for it to have a positive slope and be tangent to both the circle and the hyperbola in this way. So down here, you definitely want to take the principal root. Now with that said, this is the constraint from the hyperbola, this is the constraint from the circle. Let's set them equal to each other and solve for m, and then we'll have our slope of the line. So let's set them equal to each other. So we have the square root of 9m squared minus 4 is going to be equal to negative 4m plus 4 times the square root of m squared plus 1. So the first thing we can do, let's square both sides of this. So we get 9m squared minus 4 is equal to this term over here squared. So this is 16m squared and then plus 2 times the product of these two terms. So the product is going to be negative 16m. So it's going to be negative 32m times the square root of m squared plus 1, and then plus-- square this term-- 16 times m squared plus 1. So let's try to simplify this a little bit, get maybe the radical on one side of the equation, and see if we can simplify this even more. So a good place to start, so let's see. This term right over here is going to be 16m squared plus 16. So on the right-hand side-- let me just write it like this. So we have 9m squared minus 4 is equal to-- on the right-hand side, we have 16 plus 16. We have 32. 32m squared plus 16, minus 32m, times the square root of m squared plus 1. Now let's see. We can subtract 9m squared from both sides, so we'll subtract 9m squared from both sides. And then we can add 4 to both sides. And then we are left with-- actually, let me do it the other way, because I want to isolate this on the right-hand side. Let me do it the other way. So let's subtract 32m squared from both sides. Well, yeah, might as well do it this way. So subtract 32m squared from both sides, and then subtract 16 from both sides. The left-hand side, 9 minus 32 is going to be negative 23, because 23 plus 9, yup, is-- so this is negative 23m squared minus 20 is equal to negative 32m times the square root of m squared plus 1. Now we can square both sides of this equation. It's not the cleanest problem in the world, but hopefully, if we haven't made any mistakes, we'll get someplace productive. Actually, let's multiply both sides of this equation times negative 1 just to make it positive and simplify things. So that becomes positive, positive, and positive. If you square the left-hand side of the equation, we get 23 squared. Let me write it this way. We get 23m squared squared plus 2 times the products of this. So this would be 40-- let me just write it 2 times 20 times 23m squared, plus 20 squared, which is 400, is going to be equal to 32 squared, or I'll write it as 32m squared, times m squared plus 1. Now let's see what we can do over here. So this turns into-- this is just a big tedious problem. So I'll just write this is 23m squared squared plus-- so this is going to be 40. This is 40 times 23. So 23 times 4 is 92, so it's going to be plus 920m squared. Actually, let me expand everything out here. Let me just go straight to the numbers, so 23 times 23, 23 squared. 3 times 3 is 9. 3 times 2 is 6. 69, put a 0 here. 2 times 3 is 6. 2 times 2 is 4. So you get a 9. 6 plus 6 is 12. 1 plus 4 is 5. So you have 529m to the fourth plus 40 times 23. 4 times 20 is 80. 4 times 23 is 92. So it's going to be plus 920m squared plus 400 is equal to-- so we're going to have 32 squared. Let me figure out what 32 squared is. 32 times 32. It seems like if you're taking this exam, you might as well have your multiplication tables memorized to about 50. And I think they do that on purpose. So just to get a sense, I mean, this isn't an exam where the test takers expect everyone to get a perfect-- I think the number one score in India is on the order of 80%, which is pretty darn amazing. To be able to do roughly the math section in about an hour, to 80% right is pretty amazing. But I think to make the threshold of getting into these highly selective colleges, I think you have to get around 40% or 50% correct. Forgive me if I got that number wrong, but I think it's on that order. So they need to do that because you have 200,000 or 300,000 people in India taking it, so you need to make sure that you have a good spread of people. So this is what they do to spread people out. So we have 2 times 2 is 4. 2 times 3 is 6. And then put a 0 here. 3 times 2 is 6. 3 times 3 is 9. And for example, if I was preparing for the exam, I might even go ahead and-- you know me, I hate to memorize things, but if you had to do this type of problem on the order of five minutes, you might want to memorize these type of things ahead of time. The y-intercept for a line that intercepts a hyperbola, y-intercept for a line that intercepts a circle, who knows? But anyway. But in general, I don't think it's good to memorize in life. Because when you're actually doing math problems in your life, what matters more is you understand the underlying meaning. You normally have plenty of time to do it in your real life. These exams are kind of an artificial circumstance. But anyway, 4 plus 0 is 4. 6 plus 6 is 12. So it's 1,024, is equal to 1,024. That's the 32m squared times m squared plus 1. Or another way, this is 1,024m to the fourth plus 1,024, squared. And now we are in the home stretch. So let's subtract 529m to the fourth from both sides. So minus 529m to the fourth, this is extremely tedious. Minus 529m to the fourth. Frankly, at this point in the problem, you're better off if you just want to do it for speed, just trying out the choices they gave you and figuring out which m's and b's satisfy that. But anyway, let's just move forward-- or which m's satisfy it. That would probably be faster. But let's just solve it properly. So minus 529m to the fourth plus-- and then we also want to subtract 920m squared. So let's also subtract 920m squared. And so on this side, we're just left-- let's also subtract a 400 just to make this a proper quadratic. And also we're going to put a 400 here. So the whole left side simplifies to 0 is equal to-- 1,024 minus 529. Let's see. 1,024 minus 524 would have been 500. And so it's going to be 5 less than that. So it's going to be 495m to the fourth. Did I do that right? 1,024 minus 524 would be 500, but I'm subtracting 5 more than that. So that is going to be 495. And then 1,024 minus 920, it'll be 80 plus 24. So it's going to be plus 104m squared minus 400 is equal to 0 So we have just a straight-up quadratic equation over here. And you might not recognize it, but this is the same thing. M to the fourth is the same thing as m squared squared. So let's just solve it. So we'll get m squared. We're not solving for x anymore. We're going to get m squared using the quadratic formula. You could substitute x is equal to m squared, and this will just become a natural quadratic then. m squared is equal to negative b, negative 104, plus or minus the square root-- more fun math for us without a calculator-- plus or minus the square root of 104 squared minus 4 times a, which is 495, times c, which is negative 400. So that'll make this a positive. And you have a 400 right over here, and then all of that over 2 times 495. 2 times 495 is what? Well, it's going to be 10 less than 1,000, so it's going to be 990. So let's try to see if we can evaluate this. Actually, I'll stop this problem here. I'm already cross the 10-minute threshold. In the next video, we'll just grind through this mathematics-- It's really just arithmetic at this point-- and figure out what m is going to be equal to.