2010 IIT JEE Paper 1 Problem 50: Hyperbola eccentricity

The line 2x plus y equals 1 is tangent to the hyperbola x squared over a squared minus y squared over b squared equals 1. If this line passes through the point of intersection of the nearest directrix and the x-axis, then the eccentricity of the hyperbola is-- so let's just set this up so we know what this is even asking. Let me draw the hyperbola itself. So that is my x-axis. That is my y-axis. And this will be a left/right opening hyperbola since the x squared term here is positive. So it will open to the right like this. It'll open to the right like that. And it will open to the left like-- I could draw it a little bit neater than that. It'll open to the left like this. So that's our hyperbola right over there. And then this point of intersection right over here, if you set y is equal to 0, x would be equal to positive a. So this the point a comma 0. This right here is the point negative a comma 0. x could be plus or minus a when y is equal to 0. And just so we have some of a terminology for the hyperbola out of the way, the hyperbola will have two foci, two focus points. So let's say that this is one focus point. Let me do this in another color. This is one focus point right over here. It'll be on this major axis of symmetry for the hyperbola. And let's call that point F. Let's call that point F comma 0, where F is the x-coordinate. And then we'll have another one right over here at negative F comma 0. It'll also have two directrices. I don't know what the plural of directrix is. But these are lines, and we'll talk about the relationship between the hyperbola, the foci, and the directrices in a second. But these would be lines. One of them would sit right about here. Let me do this in another color. So you'd have a directrix right over here. And you'd have another directrix right over here. Let me draw it a little straighter than that. For this type of left/right opening hyperbola, the directrices would be vertical right over there. And the relationship between the directrix, the foci, and the eccentricity is you take any point. You take any point on the hyperbola, and the distance from that point to its closest directrix-- so for this point, it would be this directrix right here. Unfortunately, this doesn't look like a straight line, but it's supposed to be a straight line. So this distance right over here to that directrix-- so this is the distance. Let me call it the distance to the directrix. And let me call this the distance-- this is the distance to the focus on this side, the distance to the focus. The eccentricity is the ratio of this distance to that distance. So the ratio of the distance to the focus over the distance to the directrix is equal to the eccentricity. So at least now that we have the setup of everything we need to deal with. Now let's actually try to solve the problem. So they give us this line. The line is 2x plus y is equal to 1. And they say it's tangent to the hyperbola. And they say this line passes through the point of intersection of the nearest directrix and the x-axis. So it intersects the x-axis where one of the directrices intersect the x-axis. And we know where this intersects the x-axis. We can set y equal to 0. If y is equal to 0, we get 2x is equal to 1, just the y disappears. You get x is equal to 1/2. So it intersects the x-axis right over there at 1/2 comma 0. And so we know from this information right over here that this directrix must be the line x is equal to 1/2. And actually, we can draw the rest of this line right here. Its y-intercept, when x is equal to 0, y is going to be equal to 1. So that is the y-intercept. And so this line is going to look something like-- let me do it in a new color. The line will look something like this, and it will be tangent to the hyperbola. Now, our goal is to figure out this eccentricity. And to do that, I'm just going to lay out all of the formulas I know that we've proven-- and I think we've proven all of them in previous Khan Academy videos-- related to parabolas-- I'm sorry, not parabolas-- hyperbolas, eccentricity, focal point, and directrices, and then see if we can do a little bit of fancy algebra to do this. And as you know, I don't like to just regurgitate formulas. That's why we've taken some pains to prove them. But if you're doing this for especially the IIT exam, where you don't have a lot of time, it is good to have them very quickly at your disposal. So in the last video, we came up with the relationship between a tangent line and a hyperbola. We said if the tangent line has the equation y is equal to mx plus c-- and I use c here not to get confused with the b in the hyperbola-- we figured out a relationship between the m, c, the a, and the b. And the relationship we showed in the last video was c squared plus b squared is equal to m squared a squared, where, once again, the m and the c come from the line. It's the slope and the y-intercept of the line. And the a and the b come from the hyperbola. So this is one tool that seems like it might be useful. The other tool-- and I proved this. I think if you do a search on Khan Academy for hyperbola and proof, or foci, hyperbola, and proof, you get this right here in that the focal length-- this coordinate right over here, the distance from the origin to the focus point along the x-axis, or the x-coordinate of this focal point over here-- and you could actually do it for either one because you're squaring it, so it becomes positive-- it's equal to a squared plus b squared. So this seems like it might be useful in some way if we use all this information to somehow come up with the eccentricity. And then the last pieces of information that should be useful for this problem are the ones that relate the eccentricity of a hyperbola to its focal point, and a, and its directrix. And so these are-- and I won't prove them here. I think I might have proven them in previous videos. If I haven't, that's something for my to-do. But these are pretty well known. If you look it up on Wikipedia, these are well-known expressions for a hyperbola, is that the focal point is equal to the eccentricity times a, and the eccentricity is greater than 1. So this is clearly greater than a. So you multiply something larger than 1 times a, you get to the focal point and also the directrix. And in this situation, the directrix we already know is at 1/2, is at x is equal to 1/2. But I'll just write the formula here. The directrix distance, so the distance from here to the directrix, is equal to a divided by the eccentricity. And we already know, from the first part of this problem, that this is at the point x is equal to 1/2. So this is equal to 1/2. That's this distance right here. This is the line x is equal to 1/2. And this is just a known formula for hyperbolas, which I won't prove in this video because then it will just take forever. Now let's see if we can use all of this information right over here to solve for the eccentricity, to actually solve for the eccentricity. And here, it should hopefully just be a bunch of algebraic manipulation. I just want to get rid of as many variables as possible and then solve for E. So the first thing I can do is I can substitute for F in this equation right over here, and then I have an expression in terms of E, a, and b. So if I substitute for F here, F squared is going to be eccentricity squared times a squared-- I'm just squaring this over here-- is going to be equal to a squared plus b squared. And actually, let's see if we can even substitute more using this information over here. We know that a over the eccentricity is equal to 1/2. Or we could say that if we multiply both sides by the eccentricity, a is equal to 1/2 times the eccentricity. So let's substitute that into this over here. So then we get eccentricity squared times a squared. Instead of writing a squared, let's just square this thing because that's also a squared. So it's going to be times the eccentricity squared over 4-- that's just a squared; I'm just substituting it-- is equal to the eccentricity squared over 4, once again, a squared, plus b squared. And so this will simplify to the eccentricity to the fourth power over 4 is equal to the eccentricity squared over 4 plus b squared. And I don't like having these 4's in the denominator, so let's multiply both sides by 4. We get the eccentricity to the fourth power is equal to the eccentricity squared plus 4b squared. So we have an expression in terms of E and b. And so far, we used this information, this information, and this information. Oh, actually, right, and this information right over here. Now let's see if we can incorporate this somehow. And this, that relates the tangent line to the hyperbola, is useful because it actually gave us the equation of the tangent line. We actually know that this equation right here, if we write it in slope-intercept, the tangent line is y is equal to-- subtract 2x from both sides-- negative 2x plus 1. So m is negative 2. And normally, we would call this b, but I don't want to confuse it with this b. We call it c in this formula. Our y-intercept is 1. So this expression right over here, let's rewrite it. So we have c squared-- I'll do it in green-- c squared is 1 squared. So we have 1 plus-- we don't know what b squared is-- so 1 plus b squared is equal to m squared. We know that m is negative 2. Negative 2 squared is 4, 4 times a squared. Now, instead of putting an a there, we already have an expression in terms of E and b, so let's put this in terms of our eccentricity. We know that a is the same thing as 1/2 times the eccentricity, or so we can write this 4 times the eccentricity-- actually, this is an epsilon, not an E. 4 times, we just square this. Eccentricity squared over eccentricity squared over 4. I just squared this over here. And so these will cancel out, and so we have another equation, that 1 plus b squared is equal to the eccentricity squared. Or we can subtract 1 from both sides. And I want to do that so that we can substitute for b squared up in this. We really only have two formulas now or two equations, and we have two unknowns. So in theory, we should be able to solve for the eccentricity now. So we have this. Let's just solve for b squared. So we get b squared is equal to the eccentricity squared minus 1. And then we can substitute this back in for b squared and see if we can actually solve for the eccentricity. So then this expression up here is-- we have the eccentricity to the fourth is equal to the eccentricity squared plus 4 times b squared, which is this, which is the eccentricity squared minus 1. And now we have a fourth-degree polynomial in eccentricity, but let's see if we can still solve it. Hopefully, they're not going to give us something too painful. So we have the eccentricity to the fourth is equal to eccentricity squared plus 4 times the eccentricity squared minus 4. Now let's bring all of that on to the left-hand side. So we're going to subtract all of this from both sides. And we get-- all right, this is a new color. The eccentricity to the fourth power, these two over here-- actually, let me merge them. This is 1 times eccentricity squared plus 4 times eccentricity squared. So this is 5 epsilon squared or 5 times the eccentricity squared. We're going to subtract it from the other side. So this becomes minus 5 times the eccentricity squared. And then we're going to add 4 to both sides. So plus 4 is equal to 0. And now we just have to solve for eccentricity. Now, at first, this might seem complicated, but this is actually a very factorable polynomial. Think of two numbers, two factors of 4 that add up to negative 5. And hopefully, negative 4 and negative 1 pop to mind. So we can actually factor this. No quadratic formula necessary. This is factorable into epsilon squared minus 4 times epsilon squared minus 1. Negative 4 times negative 1 is positive 4. If you add the two, you get negative 5. And if this confuses you with the fourth power and the squared, you could do a substitution. So you could say-- I don't know-- some random letter. Call it some letter we haven't used yet. Call it the letter-- we already used F. Call it the letter g. You could say that g is equal to epsilon squared, and then this would become g squared minus 5g plus 4. And then this just becomes kind of a traditional factorable quadratic. But this is equal to 0. And so there's going to be two solutions over here, actually more than two solutions. So you have epsilon squared minus 4 is equal to 0 and epsilon squared minus 1 is equal to 0. This one is epsilon squared is equal to positive 4. This one is epsilon squared is equal to 1. If we were to just solve this and not think about what eccentricity is trying to tell us, we would get plus or minus 2 here, eccentricity. And we would get eccentricity is equal to plus or minus 1 here. And now the key realization is the eccentricity for a hyperbola is always greater than 1. So the only eccentricity here that is greater than 1 is positive 2. The other ones are either 1, or they're negative. So the only thing greater than 1 is positive 2. So the eccentricity of our hyperbola is positive 2. And we're done.