Watch a hyperbola break
Tonight's idea
The equation \(xy = 1\) draws one curve. Put a dial on it: \(xy = t\). Now each value of \(t\) gives its own curve, and together they make a family. The number \(t\) is called the parameter.
Every point of the plane lies on exactly one member: the point \((x, y)\) is on the member with \(t = xy\). So \((2, 3)\) is on \(xy = 6\) and on no other.
For every \(t \gt 0\) the member is a hyperbola: two separate smooth branches that never touch an axis. Turn the dial down and the branches press as close to the axes as you like. But at \(t = 0\) the curve is not a very tight hyperbola. It is a different kind of shape: two straight lines that cross. A member like that is called degenerate.
In a family, nearby members usually look alike. The values of the parameter to hunt for are the few where the shape changes kind.
Watch one: four members of \(xy = t\)
Plot. When \(t\) is not 0, no point of \(xy = t\) is on an axis: \(x = 0\) or \(y = 0\) would make \(xy\) equal 0, not \(t\). So you can divide: \(y = \tfrac{t}{x}\). At \(x = 2\) the three hyperbolas have \(y = 2\), 0.5 and 0.05.
The nearest point. For \(t = 4\), try \(x = y\): then \(x^2 = 4\), giving \((2, 2)\) and \((-2, -2)\), each at distance \(\sqrt{2^2 + 2^2} = \sqrt8 \approx 2.83\) from the origin. Exercise 1 shows no point is closer.
The member \(t = 0\). A product is 0 exactly when one of its factors is 0. So \(xy = 0\) means \(x = 0\) or \(y = 0\): every point of the \(y\)-axis together with every point of the \(x\)-axis.
What broke? For \(t = 0.1\) the curve has two pieces that do not meet, and a close zoom on any of its points shows nearly one straight line. For \(t = 0\) the curve is in one piece, and at the origin every zoom shows two lines crossing.
Your turn
- Here is why the orange dots are where they are. (a) Expand \((x - y)^2 + 2xy\) and check that it equals \(x^2 + y^2\). (b) A point \((x, y)\) is at distance \(\sqrt{x^2 + y^2}\) from the origin. Use (a) to explain why, when \(t \gt 0\), every point of \(xy = t\) is at distance at least \(\sqrt{2t}\), and exactly \(\sqrt{2t}\) only when \(x = y\). (c) Find the two nearest points, and their distance from the origin to two decimal places, for \(t = 1\) and for \(t = 0.1\).
- Turn the dial past zero, to \(t = -1\). (a) Explain why every point of \(xy = -1\) lies in the second or fourth quadrant (top left or bottom right). (b) Find the two points of \(xy = -1\) nearest the origin. The identity \(x^2 + y^2 = (x + y)^2 - 2xy\) helps. (c) Describe what you see as \(t\) goes steadily from 1 down to \(-1\).
- A different family: \(y = x^2 + t\). (a) How many times does the curve meet the \(x\)-axis when \(t = -1\), when \(t = 0\) and when \(t = 1\)? (b) For which \(t\) are there two meeting points, one, or none? (c) For \(t = -0.01\), how far apart are the two meeting points? (d) Is any member of this family degenerate?
- The family \(x^2 + y^2 = t\). (a) Describe the member for \(t = 4\), \(t = 1\) and \(t = 0.01\). (b) What is the member \(t = 0\)? (c) What is the member \(t = -1\)? (d) Both this family and \(xy = t\) do something special at \(t = 0\). How do they differ once \(t\) is negative?
Stretch: another cross
Take the family \(x^2 - y^2 = t\). (a) Factor the left side and find the member \(t = 0\). (b) Where does the member \(t = 1\) meet the axes? And \(t = -1\)? Describe the change as \(t\) passes through 0. (c) Why does no member with \(t\) not 0 ever touch the member \(t = 0\)? (d) Take a point on the member \(t = 0\) and multiply both its coordinates by any number \(k\) that is not 0. Show the new point is still on that member. What does this tell you about zooming in on the origin? Compare with zooming in on \((1, 1)\).
Answers
- (a) \((x - y)^2 + 2xy = x^2 - 2xy + y^2 + 2xy = x^2 + y^2\). (b) On the curve \(xy = t\), so \(x^2 + y^2 = (x - y)^2 + 2t\). A square is never negative, so this is at least \(2t\), and it equals \(2t\) only when \(x - y = 0\). Take square roots: the distance is at least \(\sqrt{2t}\). When \(x = y\), \(x^2 = t\), so the nearest points are \((\sqrt t, \sqrt t)\) and \((-\sqrt t, -\sqrt t)\). (c) \(t = 1\): \((1, 1)\) and \((-1, -1)\), distance \(\sqrt2 \approx 1.41\). \(t = 0.1\): about \((0.32, 0.32)\) and \((-0.32, -0.32)\), distance \(\sqrt{0.2} \approx 0.45\). The corner of each branch slides toward the origin and would arrive at \(t = 0\).
- (a) Neither \(x\) nor \(y\) is 0, and a negative product needs one positive factor and one negative. \(x \lt 0, y \gt 0\) is the second quadrant and \(x \gt 0, y \lt 0\) is the fourth. (b) On the curve, \(x^2 + y^2 = (x + y)^2 + 2\), which is at least 2 and equals 2 only when \(y = -x\). Then \(-x^2 = -1\), so the points are \((1, -1)\) and \((-1, 1)\), at distance \(\sqrt2 \approx 1.41\). (c) Two branches, top right and bottom left, press into the axes. At \(t = 0\) they are the axes. Then two branches peel away into the top left and bottom right. The member \(xy = -1\) is the mirror image of \(xy = 1\) in the \(y\)-axis. Only at \(t = 0\) is the curve in one piece.
- (a) On the \(x\)-axis \(y = 0\), so \(x^2 = -t\). For \(t = -1\): \(x = 1\) or \(-1\), two points. For \(t = 0\): only \(x = 0\), one point. For \(t = 1\): \(x^2 = -1\) has no real solution, no points. (b) Two when \(t \lt 0\), at \(x = \pm\sqrt{-t}\). One when \(t = 0\). None when \(t \gt 0\). (c) \(x = \pm 0.1\), so 0.2 apart. The two points merge as \(t\) rises to 0. (d) No. Every member is the same parabola \(y = x^2\) slid up or down by \(t\), so no member changes kind. What changes at \(t = 0\) is how the curve meets the \(x\)-axis: it only touches. A special value of the parameter does not have to mean a degenerate member.
- (a) Circles centred at the origin with radius \(\sqrt t\): radius 2, 1 and 0.1. (b) Squares are never negative, so the sum is 0 only when both are 0. The member is the single point \((0, 0)\): a circle shrunk to nothing, and degenerate. (c) The left side is never negative, so no real point satisfies it. The member is empty. (d) The hyperbolas carry on past 0, with branches in the other two quadrants. The circles do not: for negative \(t\) there are no real points at all.
Stretch. (a) \((x - y)(x + y) = 0\) means \(y = x\) or \(y = -x\): two lines crossing at the origin. (b) \(t = 1\): putting \(y = 0\) gives \(x = \pm 1\), and putting \(x = 0\) gives \(-y^2 = 1\), impossible. So it meets only the \(x\)-axis, at \((1, 0)\) and \((-1, 0)\), and its branches open left and right. \(t = -1\) meets only the \(y\)-axis, at \((0, 1)\) and \((0, -1)\), opening up and down. As \(t\) passes through 0 the branches close onto the cross and reopen in the other two gaps. (c) At every point of the two lines \(x^2 - y^2 = 0\), which is not \(t\). The cross is the pair of lines that the other members approach far from the origin without touching: their asymptotes. (d) \((kx)^2 - (ky)^2 = k^2(x^2 - y^2) = 0\). So scaling the picture about the origin leaves this member unchanged: at every zoom you see the same two crossing lines, never one. Near \((1, 1)\) the line \(y = -x\) is \(\sqrt2 \approx 1.41\) away, so a close zoom shows only the single line \(y = x\). The origin is the one point of this member that never looks like a single line. It is called a singular point. One more thing: with \(u = x - y\) and \(v = x + y\) the equation reads \(uv = t\). It is the first family again, drawn on diagonal axes.