After going through many of these Indian
Institutes of Technology Joint Entrance Exam problems,
I realize that there are a lot of problems where they really
just expect you to know something.
So that's what I'm going to cover in this video, one
of those things that they just expect you to know.
What we're going to do is come up
with the relationship between a conic section--
and a particular hyperbola is what we'll focus on--
and a tangent line.
We've done this in a previous problem.
But that wasn't in the general case.
So let's say we have a left right opening hyperbola.
So it'll have the equation, x squared over a squared minus
y squared over b squared is going to be equal to 1.
And so if I were to draw that hyperbola
it would look something like this.
That's the x-axis.
That's the y-axis.
And then it opens to the right.
I could draw a better bottom half.
It opens to the right.
And it opens to the left.
And in case you're curious, this point right
here, if you set y is equal to 0,
this point right here is a comma 0.
And this right here is negative a comma 0.
So what I want to do is figure out
a relationship between these a's and b's and the equation
of a tangent line.
So let's say I have a tangent line that
looks something like this.
Let's say it's tangent only at that point right over there.
And so it would look like-- let me draw it a little bit better
than that-- it would look something like that.
And let's say the equation for this tangent line
is y is equal to mx, where m is the slope,
plus-- instead of saying b for the y-intercept.
So normally, we would call the y-intercept b for a line.
We've already used the b here in the equation for the hyperbola.
So let me just call this c.
So the c-- this is a little unconventional.
This is going to be the y-intercept.
So let's see if we can come up with a relation between
the m's, the c's, the a's, and the b's.
And we already used it in one of the IIT problems.
And I suspect that the next one I'm going to do
will also use this.
If you all have seen a lot of Khan Academy videos,
you know that I always like to prove things
from first principles, because in life, you can't just
memorize formulas.
You won't know where they came from.
You'll memorize them wrong.
You won't understand what they actually mean.
But if you're going to take the IIT JEE exam,
I would recommend that you-- I have an appreciation
for how little time they give these problems.
And if you have to prove from first principles,
you won't be able to prove them.
So let's just come up with the relation.
Let me stop talking without drawing.
So let's just see where they intersect.
And the whole insight here is that they're only
going to intersect in one point.
So what I'm going to do here is solve for y squared.
So over here, we can multiply both sides of this equation
by-- let's multiply both sides by negative b squared.
So you get negative b squared over a squared
x squared plus y squared-- I multiply by negative
b squared-- is equal to negative b squared.
And now, let's add this thing to both sides of this equation.
And we get y squared is equal to b squared
over a squared x squared minus b squared.
So I just rewrote the equation for the hyperbola.
And let's also write this in terms of y squared.
And then we can set them equal to each other.
So over here in this greenish yellow-like color,
if we square both sides, we get y squared
is equal to m squared x squared plus 2 times the product
of those both terms.
So plus 2 mcx plus c squared.
So in order for them to intersect,
they both have to be at the same place at some x and y.
So we can set this y squared being equal to that y squared,
and then try to solve for the x.
Obviously, we won't be able to solve for the x,
because there's so many variables.
We can find a relationship between this a, this b, this m,
and this c, so there's only one point of intersection,
which by definition, it would have
to be at the tangent point.
So let's do that.
So we have m squared x squared plus 2 mcx plus c squared
is equal to b squared over a squared x squared
minus b squared.
And I've done a very similar exercise
to this in a previous IIT JEE video.
But here, I just want to focus just on the most general case
so that we have something that we can add to our tool kit.
So let's write this in terms of a quadratic equation in x.
So if we subtract this from both sides, we get m squared.
We have m squared minus b squared
over a squared x squared.
That's that term and that term, right over there.
And then my only just first degree x term
is right over here.
So plus 2mcx-- I'll write it in-- plus 2mcx.
And then finally, plus, we have a c squared,
and then we're actually going to have
a plus another b squared over there.
So this is going to be plus c squared plus b squared.
Now, in order for this equation to only have one solution--
let me write it down.
This is going to be equal to 0.
In order for this thing to only have one solution,
the discriminant of this quadratic equation--
remember, when you do the quadratic equation-- and these
are completely different, so negative b
plus or minus the quadratic formula b squared minus
4ac over 2a-- you're only going to have one
solution if this thing over here,
if the discriminant over there is equal to 0.
If b squared minus 4ac is equal to 0,
then you only have the solution negative b over 2a.
So in this situation, for the tangent line,
you can only have one solution, one x
that satisfies this equation.
So b squared minus 4ac is going to have
to equal 0 in the quadratic formula.
These are different b's and a's and c's
than the ones we're using here.
Over here, our b is that right over there,
the coefficient on the x term.
So that squared is 4m squared c squared.
And we're going to subtract from that minus 4 times a.
a is all of this business right over here.
And just to simplify things, let me make this plus,
and then multiply this times a negative sign.
And it'll reverse the sign.
So I'll put minus 4-- I'll put a plus here-- times,
instead of writing, since I have to multiply this by negative 1,
it'll now be b squared over a squared minus m squared.
And then the c is this term right over here.
This right here is c, 4ac.
c squared plus b squared.
And this thing is going to equal 0 if this line is tangent,
if we only have one solution.
So the first thing that we can do to simplify this
is, we can divide both sides of this equation by 4.
And if we do that, this becomes--
I wanted to do that in black-- this becomes a 1.
So we can ignore that.
And then this becomes a 1.
So that simplified our equation a good bit.
And now, let's multiply the second part right over here.
So we have b squared over a squared times c squared.
That is b squared c squared over a squared.
And then b squared a squared times
b squared-- so plus b to the fourth over a squared.
Then you have negative m squared times c squared.
So let me do this in another color.
So then you have for the m squared,
negative m squared c squared.
And then you have negative m squared b squared.
And of course, this is all going to be equal to 0.
And we have this m squared c squared out front over here.
And lucky for us, this cancels with this.
And what do we have left?
Now, every term here is divisible by b squared.
So let's divide every term by b squared.
So this will just become a 1.
This will become b to the second power.
And then this will just become a 1.
And then let's multiply everything by a squared,
just so we get rid of the fractions.
So when you multiply everything by a squared,
this term right over here becomes a c squared.
This term right over here is just a b squared.
And then all we have left, and then we
have this negative m squared.
Remember, we're multiplying by a squared.
So minus a squared m squared is equal to 0.
Or we could add this to both sides of the equation.
We get c squared plus b squared is
equal to a squared m squared.
And what's really neat about this is,
we now have a very simple relationship.
If we know the equation of the line right over here,
we would then, if we know what m and c are,
we then have an interesting relationship for a and b.
If we know what a and b are, we have
an interesting relationship for the equation of the line.
And maybe if we have a few other constraints,
we could actually solve for them.
But we'll actually take this and use
this in the next IIT problem we're going to do.