Locus and Change of Axes

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Picture a ladder leaning against a wall. It begins to slip: the top slides down the wall, and the foot slides along the floor. Now watch only the middle of the ladder. What path does it trace? Is it a straight line, or some kind of curve? Hold that guess. Today we learn how to find the path of a moving point, called its locus, and how to shift and turn the axes to make an equation simpler. Both ideas come up often in JEE coordinate geometry. So, let's begin.

0:35Locus: the path of a moving point

So, what is a locus? When a point moves, but always obeys some condition, the path it traces is called its locus. Okay? Now the method. Step one: call the moving point h comma k, in round brackets. Fresh letters, h and k, keep the moving point separate from the other x's and y's in the question. Right? Step two: translate the given condition into an equation connecting h and k. Clear? Step three: eliminate any parameter. Sometimes the point depends on an extra quantity, like t, that changes as the point moves; that quantity is called a parameter. And at the very end, replace h comma k by x comma y. Many students forget this last step, so don't! Got it? Let us see it once. The point P is t squared comma two t; the small two up top means squared. So h equals t squared, and k equals two times t. Now tell me, how do we get rid of t? Square k, yes, exactly! k squared is four times t squared, and t squared is just h, so k squared equals four times h. The double arrow means, gives. Replace h and k with x and y, and we get y squared equals four times x, a parabola. The parameter t has completely gone. With me?

1:57Shift of origin

Now our second idea, change of axes, and first, the shift of origin. We slide both axes, without turning them, so that the origin moves to a new point, alpha comma beta. Alpha and beta are Greek letters, here just two fixed numbers. Okay? Remember, the point itself does not move; only its coordinates change. Small x and small y are the old coordinates, and capital X and capital Y the new ones. Right? So let me ask you this. The new origin has new coordinates zero comma zero. What are its old coordinates? Alpha comma beta, exactly! So the old coordinates are the new ones plus the shift: small x equals capital X plus alpha, and small y equals capital Y plus beta. Clear? Now a quick test of signs. Suppose the new origin is two comma minus three. Is small y equal to capital Y plus three, or capital Y minus three? It is capital Y minus three, correct! Here beta is minus three, and adding minus three means subtracting three. So small x equals capital X plus two, and small y equals capital Y minus three. Students often flip this sign, so be careful. Got it?

3:13Rotation of axes through an angle theta

Next, rotation of axes. The origin stays fixed, and we turn both axes anticlockwise through an angle theta, the Greek letter on screen. For a clockwise turn, just use minus theta. Okay? Small x equals capital X times cos theta, minus capital Y times sine theta. Notice the minus sign; it lives in the x formula. Right? Next, small y equals capital X times sine theta, plus capital Y times cos theta. Here, both terms are plus. Clear? Let us test these formulas. Take the point one unit along the new X axis, so capital X is one and capital Y is zero. What are its old coordinates? Cos theta comma sine theta, yes! That is the point one unit from the origin, at an angle theta from the old x axis, exactly where the turned X axis points. So the formulas make sense. With me? Now, here is a trap many students fall into. Both formulas give the old coordinates in terms of the new ones. So to change an equation, substitute them straight into the old equation in small x and small y, and do not mix up the direction. Okay?

4:31Example 1: locus of mid-point of a sliding rod of length 10

Now example one, and it is our ladder puzzle in coordinates. Take the floor as the x axis and the wall as the y axis. Here, a rod of length ten slides with its ends on the coordinate axes. Find the locus of its mid-point. Pause the video and try it first. Done? Okay, let's check. Call the mid-point h comma k. One end is on the x axis, and the other is on the y axis. So the ends are at h comma zero and zero comma k, right? No, no, careful, that is the classic slip here! The mid-point of those two would be h over two, comma, k over two. The mid-point averages the ends, so the ends must be at two times h comma zero, and zero comma two times k. Clear? Now use the length. The ends are ten apart, so by the distance formula, two times h, the whole squared, plus two times k, the whole squared, equals ten squared. Okay? That gives four times h squared, plus four times k squared, equals one hundred. Take out the common four: h squared plus k squared equals one hundred over four, which is twenty five. Right? Finally, replace h comma k by x comma y. The locus is x squared plus y squared equals twenty five: a circle with its centre at the origin, and radius five, half the rod. So our puzzle is solved: the middle of the ladder moves along an arc of this circle, not a straight line. Nice, isn't it?

6:03Example 2: rotate the axes by 45 degrees in xy = 2

Now a harder one, example two. Rotate the axes through forty five degrees in the curve x y equals two. The small circle after forty five means degrees. Pause the video and try it first. Done? Okay, let's check. We use the rotation formulas with theta equal to forty five degrees. So first, what are cos forty five degrees and sine forty five degrees? Both are one over root two, correct! The tick-shaped sign is the square root sign. Right? So small x equals capital X minus capital Y, over root two. In the same way, small y equals capital X plus capital Y, over root two. Clear? Now multiply small x and small y. Look at the tops first: what is capital X minus capital Y, times capital X plus capital Y? It is capital X squared minus capital Y squared, yes, the difference of two squares! And on the bottom, root two times root two is two. So x y equals capital X squared minus capital Y squared, all over two. Got it? The curve says this equals two, so double both sides: capital X squared minus capital Y squared equals four. That is our answer. With me? And look what happened. The product term x y has completely gone. Same curve, seen from turned axes, in a much simpler form. You will meet it again in the hyperbola chapter, as the rectangular hyperbola. That is the power of rotating the axes!

7:44Quick revision: locus steps, shift and rotation formulas

So, quick revision. For a locus, take h comma k, write the condition, eliminate the parameter, then put x comma y. For a shift of origin to alpha comma beta, small x equals capital X plus alpha, and small y equals capital Y plus beta, okay? And for an anticlockwise rotation through theta, small x equals capital X cos theta minus capital Y sine theta, and small y equals capital X sine theta plus capital Y cos theta. Now go and try the practice problems on this page. All the best!