now that we've laid a foundation and understand a bit more about tangent lines and how to build formulas for them we can use that to answer some interesting geometric problems little puzzles like the one that follows we're asked here to find the equations of all the lines through the origin that are also tangent to the parabola given here the key thing we're going to need for this is that we're talking about tangent lines that's the same or we'll have the same formula as linear approximations that we've seen previously in this kind of problem having a sketch especially one that we can label with some details can really help so here's a rough sketch of a parabola that opens upwards that's been shifted away from the origin and what we're interested in are the points where the lines through the origin are also tangent to the parabola so we can imagine right here there might be a point here where that happens but it's not going to be every point for example if we pick this point on the parabola and draw a tangent line there it definitely is not going through the origin it's shooting off to other places likewise if we drew a tangent line at that location it's also not going to go through the origin so what we'd expect to see is maybe two different points so we might see it looks like two points maybe one point where the tangent line goes through the origin now how are we going to actually solve for those locations what makes them special well we're going to need a few different things and let's just draw one of these as a hypothetical let's consider this point here it'll be special let's say we're looking for that point and based on the nomenclature of our earlier formulas let's call that point x equals a but a is the x value we're going to need to keep x free and this is where it can get a bit confusing for some students we're going to need to keep x free because we want the equation of this line equation of the tangent line well we know how this equation works it's y equals the slope and here we have to be very careful about the x's versus the a's if we pick a point a and we draw the tangent line that's the the a value that goes into our slope calculation then we multiply that times x minus a that's another key ingredient here this is the constant and x is the point that we're considering along the line so x is still a variable and last but not least plus f of a so this is the equation of a tangent line based on the point x equals a on the parabola fair enough in fact that actually defines the green lines as well if you picked a to be minus three say in the sketch you would get the formula for this green line you've picked one it might be this green line so how does that help us well we want the a's such that the line passes through the origin and the origin specifically has a coordinate 0 0 so a is the anchor point for the line but what we want to guarantee is that x equals 0 y equals 0 satisfies the line equation this is x changes as we go to left to right and the y changes correspondingly i want to know if i plug in x equals zero well i also get a height of zero that's not true in this line here and this line if we went to x equals zero we would get a y that's positive on this one to follow it all the way down where x equals zero it would give us something massive and negative so this is going to be a special condition on the line and so we can plug that in we want 0 copying this over to equal f prime of a times x minus a plus f of a and actually the x is also now a zero all right well this so far hasn't used the details of the parabola that we were given well that's exactly what we have for f of a and we can find out the value for f prime of a so this function gives us the y values or if you like this whole thing here is f of x and so what we can do is plug those values in we'll do that on the next page with a 0 subbed in if our function is this then we can take the derivative and find the slope at any point and once we've done those two calculations or restated this formula and done the derivative calculation then we can evaluate both of them at a which is almost too simple substitution we just replace all the x's with a's and then we sub those into our tangent line formula which satisfies zero zero and that was 0 equals our slope at a 2 a minus 2 times our coordinate x of 0 minus a plus f of a so a squared minus 2a plus 4. so going back to the picture we wanted the point a where if we were to build a line that's this formula here it would satisfy 0 0 as x and y coordinates on that line and the height we would have for our a comes from the parabola so we use the parabola formula there and the slope we would use for that line comes from the derivative of the parabola to get the right slope and this gives us an equation in a where the only variable left to solve for is that a and when it pops out it should be the location on the parabola that we're looking for so let's tidy this up and solve for a we'll have zero equals minus two a squared plus two a's plus an a squared minus two a plus four together this gives us negative a squared plus four that's nice because that gives us a squared equals four or a equals two and minus two and those would be the points on the parabola specifically the x coordinates where the tangent line will pass through the origin this is practically crying out for a visual representation to confirm our answer we're going to do that on the next page the parabola we're working with is restated here and we can do a quick table of values to get a sketch of this we know we want to go from around negative 2 certainly up to 2. let's go up to 3 and see what happens we plug in the coordinates and get the corresponding y values we're going to get 12 7 4 then it'll be down to three then back up to four and there we go we're actually back up to seven for that and if we draw that out on our axis here everything's above zero so we'll center our axis there and we will sketch out our coordinates x and y there's exactly 12 steps in this grid so if we use our scaling like so and make this our six mark then we can easily get the rest of the graph sketched in here four three four and seven sketching that parabola and now we take the two points that we found notice it's not symmetric around zero that's fine that wasn't quite what we were asked for but if we go up to this point here and draw the tangent line lo and behold it looks like that tangent line does in fact pass through the origin we draw the tangent line there it ends up through the origin likewise with a different slope but still satisfying that same property we can see that at least to a rough sketch the tangent line that passes through that point also goes through the origin that is not the case for other points if we were to pick this point and draw the tangent line very clearly it does not go through the origin at all it ends up over here somewhere so these two points at x equals two and x equals minus two those tangent lines to the parabola pass through the origin this is a little less convincing than the algebraic version but it certainly gives us some reassurance that what we found for the algebra is correct