If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

# Representing a line tangent to a hyperbola

## Video transcript

after going through many of these Indian Institute of Technology joint entrance exam problems I realize that there are a lot of problems but 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 and what we're going to do is come up with the relationship between a conic section in particular a 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 just have 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 one 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 can draw a better bottom half it opens to the right and it opens it opens to the left and in case you're curious this point right here if you said Y is equal to zero this point right here is a comma zero and this right here is negative a negative a comma zero so what I want to do is figure out a relationship between these a 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 a little bit better than that it would look something something like that and let's say the equation for this tangent line is y is equal to M X where m is the slope plus instead of saying B for the y-intercept so normally we would call the y-intercept B for line we've already used the B here in the equation for the hyperbola so let me just call this C so the C 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 MS the C's and the A's and the B's and we can actually this 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 and as if y'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 je exam I would recommend this you because I have appreciation for how little time they give these problems and if you have to prove from first principles you won't be able to approve them so let's just come up with the relation let me stop talking or 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 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 and so you get negative b squared over a squared x squared plus y squared right I multiplied it by negative B squared is equal to negative 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 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 2m C X plus 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 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 M Squared x squared plus 2 see X plus C squared is equal to is equal to V squared over a squared x squared minus B squared and I've done a very similar exercise to this in a previous a I tje 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 toolkit so let's write this in terms of a quadratic equation in X so if we if we subtract this from both sides we get M squared X we get M Squared we have M squared minus B squared over a squared minus B squared over a squared x squared x squared so I just 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 2m C X I'll write it in plus 2m C X and then finally plus plus we have a C squared and then we're actually going to have a plus a another V squared over there so this is going to be this is going to be plus C squared plus C squared plus 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 is going to be equal to 0 in order for this thing to only have one solution the discriminant of this quadratic equation remember if I 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 are only going to have one solution if this thing over here 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 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 Ones we're using here over here RB our B is that right over there the coefficient on the X term so that squared is 4 M Squared C squared and we're going to subtract from that minus 4 times a a is a is all of this business right over here and just to simplify things let me make this plus and then multiply this times the negative sign and it will 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 will 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 if we only have one solution so let's see the first thing that we can do to simplify this is we can divide both sides of this equation 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's 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 and then you have negative m squared times C squared so let me do this in another color so then you have negative 4 the M Squared negative M Squared C squared and then you have negative M Squared B squared 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 M Squared C squared out here and lucky for us this cancels with this and what do we have left now every term here has 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 one and this 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 C squared this term right over here is just a B squared and then and then all we have left and then we have this negative M squared we're multiplying by a squared so minus a squared M Squared is equal to zero or we could add this to both sides of this equation we get C squared plus B squared is equal to is equal to negative 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 line if we know the line we have a 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 IT problem we're going to do