Part C: Concept Checks

Transcript

0:00Introduction

Hello everyone! Let us start with a puzzle. Take a circle and a straight line. If the line meets the circle at exactly one point, the line is a tangent; it just touches. Now take a hyperbola instead. Can a line meet a hyperbola at exactly one point, and still not be a tangent? Many students say no, that is impossible. Hold that thought. Today is Part C of our coordinate geometry problem bank: quick concept checks, the kind of ideas JEE can test in statement form. So, let's begin.

0:35When is a second-degree equation a circle?

So, our first check: when does a second degree equation represent a circle? Here is the general one: a times x squared, plus two h times x y, plus b times y squared, plus two g x, plus two f y, plus c, equals zero. The small two up top means squared. Notice the twos in front of h, g and f. Okay? Think of any circle equation from chapter three. What must be true of a, b and h? Yes, exactly! The coefficients of x squared and y squared must be equal, and not zero; the slashed equals sign means not equal. And h must be zero, so there is no x y term at all. Got it? Next, divide by a, so it starts with x squared plus y squared, and call the new coefficients g, f and c again. From chapter three, the centre is minus g, comma, minus f, in round brackets, and the radius is the square root of, g squared plus f squared minus c. With me? So, equal coefficients and no x y term, and we always get a circle, right? No, no, careful, because there is a third condition! The radius is a square root, so g squared plus f squared minus c must be greater than or equal to zero; that is the greater than sign with a line under it. If it equals zero, the radius is zero: a point circle. If it is negative, there is no real circle at all. This trap comes up often. Clear?

2:13Which conic? Check Δ, then compare h² with ab

Now the next check: which conic does a second degree equation give? First compute capital delta, the Greek letter that looks like a little triangle. Delta equals a times b times c, plus two f g h, minus a times f squared, minus b times g squared, minus c times h squared. Okay? Now tell me, why should we find delta before anything else? Because the rule works only when delta is not zero. Then the conic is non-degenerate, a genuine conic, not a broken-down case. Right? Then compare h squared with a times b. If they are equal, the curve is a parabola. If h squared is less than a times b, with the less than sign, it is an ellipse. And if h squared is greater, with the greater than sign, it is a hyperbola. Got it?

3:03One common point: always a tangent?

Now, back to our puzzle. On the right is the hyperbola, x squared over a squared, minus y squared over b squared, equals one. The red dashed lines are the directrices, S and S prime are the foci, the little dash read as prime, and P is a point joined to both. Today we need only the grey dashed lines, the asymptotes: y equals plus or minus b over a, times x; that is the plus-minus sign. The branches come closer and closer to them, but never meet them. With me? Take a line parallel to an asymptote, shifted a little. How many times does it meet the hyperbola? Such a line is y equals b over a, times x, plus c, with c not zero. Same slope, so it is parallel to one asymptote. Okay? Substitute it into the hyperbola. The x squared terms cancel, leaving a linear equation in x, with exactly one root. So the line meets the hyperbola at exactly one point. Got it? One common point, so it must be a tangent, just like for the circle, right? Oh no, careful here, and check the condition of tangency. From chapter six, the line y equals m x plus c touches this hyperbola only if c squared equals a squared times m squared, minus b squared. Clear? Now put m equal to plus or minus b over a, the slope of an asymptote. Then a squared times m squared is b squared, so c squared is zero, and c is zero. But that line is the asymptote itself, which meets the hyperbola nowhere. So no tangent is parallel to an asymptote. Got it? So the answer to our puzzle is yes! Any line parallel to an asymptote, other than the asymptote itself, meets the hyperbola exactly once, but it is not a tangent; it crosses the curve.

5:00Example 1: can two different circles share three points?

Clear? Now Example one. Can two different circles have three common points? Pause the video and try it first. Done? Okay, let's check. Suppose two different circles do share three points. Three points either lie on one line, or they do not: two cases, okay? First case: all three points on one line. Is that possible for points of a circle? No, it is not possible. From chapter three, any line cuts a circle at two points, touches it at one, or misses it. So no line meets a circle in more than two points. Clear? Second case: the three points are not on one line; we call them non-collinear. And three non-collinear points fix a unique circle: exactly one circle passes through them. With me? But both our circles pass through these three points, so both must be that one circle: the same circle! That breaks our assumption that they are different. Right? So the answer is no: two different circles share at most two points. Got it?

6:08Example 2: classify x2 + 2xy + y2 - 4x + 4 = 0 (a parabola)

Now Example two, a little harder. Classify x squared, plus two x y, plus y squared, minus four x, plus four, equals zero. Pause the video and try it first. Done? Okay, let's check. First, read off the coefficients. The x y term has coefficient two, so h is two, right? No, no, wait, because in the general equation the x y term is two h times x y! So here, a is one, two h is two, b is one, two g is minus four, two f is zero, and c is four. Okay? Now halve them: h is one, g is minus two, and f is zero. Taking h as two is the classic slip. Clear? Now compare. h squared is one, and a times b is one times one, which is also one. So h squared equals a times b. Can we call it a parabola already? Not yet, because we must check delta first! Since f is zero, the terms two f g h and a times f squared both vanish. So delta is a times b times c, minus b times g squared, minus c times h squared. Each of these three terms is four, so delta is four minus four minus four, which is minus four, not zero. With me? So the conic is non-degenerate, and h squared equals a times b: the curve is a parabola. Here is a topper's habit: halve the coefficients first, check delta, then compare h squared with a times b. Got it?

7:51Revision: circle test, conic type, asymptotes, two circles

So, quick revision. For a circle: equal coefficients, not zero, no x y term, and after dividing by a, g squared plus f squared minus c at least zero. To classify a conic, make sure delta is not zero, then compare h squared with a times b: equal, parabola; less, ellipse; greater, hyperbola. Any line parallel to an asymptote, other than the asymptote itself, meets the hyperbola once, but it is not a tangent. And two different circles share at most two points. You handled four tricky ideas today, very well done! Now try the practice problem on this page. All the best!