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## IIT JEE

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# 2010 IIT JEE Paper 1 Problem 50: HyperbolaÂ eccentricity

## Video transcript

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.