Overview

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Suppose a straight line meets a curve at exactly one point. Then the line must be a tangent to that curve, isn't it? Most students say yes, without a second thought. Hold on to your answer; by the end of this video, you will know the truth. This chapter is the problem bank for the whole of coordinate geometry, and that puzzle is one of its concept checks. So let us see how to use this chapter, and solve two of its problems together, so let's begin!

0:33How to use this chapter

So, let's open this chapter. Part A has six solved problems, at JEE Main level. Okay? And look at the spread: one each on the centroid, a line, a circle, the parabola, the ellipse and the hyperbola. Six topics, one problem from each. Right? Now, how should you use them? The course says: attempt Part A as a thirty minute test, after Chapters one to seven; the short dash means to. Clear? So, a quick question for you. Six problems in thirty minutes. How much time can you give each problem? Thirty over six gives five. So five minutes a problem, with a clock running, just like the real exam. Got it? Then Part B: five longer problems, at JEE Advanced level. Longer problems, so expect more steps in each one. With me? And Part C: four concept checks. They test the ideas that are often asked in statement form, where you decide whether a claim holds, and say why. Okay?

1:39Part C: a statement needs a reason

Now, two statements from Part C, the concept checks. Here is the first one: can two different circles have three common points? The answer is no, and here is the reason. If the three points are not collinear, that is, not on one straight line, they fix a unique circle. So the two circles would be the same circle. Okay? And if the three points are collinear? Then no circle can pass through all three, because a line meets a circle in at most two points. Both cases covered, so two different circles share at most two points. Right? Now, before the second statement, let us recall the asymptotes of a hyperbola. For x squared over a squared, minus y squared over b squared, equals one, the asymptotes are y equals plus or minus b over a, times x. The small two up top means squared, and the plus-minus sign gives two lines, one for each sign. With me? Now the second statement, which is our puzzle. Can a line meet a hyperbola at exactly one point, without touching it? One common point, so it must be a tangent, isn't it? Now the answer: yes, it can! Any line parallel to an asymptote, other than the asymptote itself, meets the hyperbola exactly once, and still does not touch it. So our puzzle is solved: one common point does not always mean a tangent. In Part C, the trap is a quick yes from habit, so check every case. Got it?

3:11Example 1: eccentricity of ellipse 16x2 + 25y2 = 400

Now, Example one, from Part A, marked easy. Find the eccentricity of sixteen x squared plus twenty five y squared equals four hundred. Remember, five minutes. Pause the video and try it first. Done? Okay, let's check. Many students glance at this and say: the bigger number, twenty five, goes with y squared, so the major axis is along y, right? No, no, careful, that is the trap! Always bring it to standard form first: divide both sides by four hundred. Sixteen over four hundred is one over twenty five, and twenty five over four hundred is one over sixteen. So x squared over twenty five, plus y squared over sixteen, equals one. With me? Now the bigger number, twenty five, sits under x squared. So a squared is twenty five, and b squared is sixteen. Then a is five, which is greater than b, which is four; the angle sign means greater than. So the major axis is along x, the opposite of the quick guess. Clear? Now, for this ellipse, which formula links a, b and the eccentricity e? Yes, exactly: b squared equals a squared times, one minus e squared. Put in the numbers: sixteen equals twenty five times, one minus e squared. Okay? Divide by twenty five: one minus e squared is sixteen over twenty five, so e squared is nine over twenty five. So e is three over five. For an ellipse, e must lie between zero and one, and three over five does, so that is our answer. Right?

4:54Example 2: locus of P, tangents to y2=4ax at theta, 90-theta

Now a harder one, from Part B, marked hard. Tangents from a point P to y squared equals four a x make angles theta and, pi over two minus theta, with the x axis. Theta is a Greek letter standing for an angle, and pi over two is ninety degrees. Find the locus of P, the path it traces. Pause the video and try it first. Done? Okay, let's check. Theta plus, pi over two minus theta, is pi over two. So the tangents are perpendicular, and perpendicular tangents meet on the directrix, x equals minus a, right? No, no, wait, that is the trap! These angles add up to pi over two. For perpendicular lines, the angles they make with the x axis must differ by pi over two, not add up to it. So let us not guess; let us work it out. Okay? Here is the tool from the parabola chapter. The line y equals m x plus c touches y squared equals four a x if and only if c equals a over m. So the tangent of slope m is y equals m x, plus a over m. Right? Let P be h comma k. The tangent passes through P, so k equals m times h, plus a over m. Multiply through by m; the double arrow means which gives. We get h times m squared, minus k times m, plus a, equals zero. Clear? Now this is a quadratic in m. So here is a question for you: what is the product of its two roots? Yes, exactly, the constant term over the leading coefficient! And its two roots are the slopes of the two tangents drawn from P to the parabola. So m sub one times m sub two equals a over h; the small numbers below are subscripts, naming the two slopes. With me? Now use the angles. m sub one is tan theta, and m sub two is tan of, pi over two minus theta, which is cot theta. And tan theta times cot theta is one. So the product is plus one, not minus one, which is why these tangents are not perpendicular. Got it? So, a over h equals one, which gives h equals a. Replace h by x, and the locus of P is the line x equals a, the line of the latus rectum. So it is not the directrix at all; read the angles carefully, okay?

7:28Quick revision: test plan, one-point trap, m1m2 = 1 not -1

So, quick revision. Part A is a thirty minute test, Part B is at JEE Advanced level, and Part C tests statements. One common point with a hyperbola need not mean a tangent. And go to standard form first; and when two angles add up to pi over two, the slopes multiply to plus one, not minus one. Now open Part A, set a thirty minute clock, and start solving. All the best!