Definition, Standard Form and Asymptotes

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. For an ellipse, the sum of the distances from two fixed pins stays the same. Now keep the difference of the two distances fixed instead. What curve do we get now? Does it close up, like the ellipse? The first figure will answer that. Hyperbola questions come up often in JEE papers, so let's begin.

0:25Two ways to define a hyperbola

So, look at the figure: two separate branches, in dark blue. That is a hyperbola, and it does not close up at all! The first definition is the locus of a point P with P S equal to e times P M; the small dot means times. The greater than sign says e is more than one. Okay? Here S is the focus, and P M is the distance of P from the directrix; the dashed red lines in the figure are directrices. Right? Now the second definition. Take two fixed points, S and S prime; the little dash is read as prime. These are the foci. The orange lines are S prime P and S P, and the two vertical bars mean modulus: their difference, made positive, always equals two a. Clear? Now let me ask you this. For the ellipse, the pins had to be closer than two a. Here, must S S prime be more than two a, or less? Yes, exactly, it must be more! In the triangle SPS prime, one side minus another is always less than the third side. So two a is less than S S prime. Got it?

1:37The standard form

Now the standard form, centred at the origin. The course writes x squared over a squared, minus y squared over b squared, equals one; the small two means squared. And b squared equals a squared, times e squared minus one. With me? Now tell me, what is e squared in terms of a and b? The ellipse habit says one minus b squared over a squared, right? No, no, careful, that formula belongs to the ellipse! Divide by a squared: e squared is one plus b squared over a squared. Plus, not minus, and that is why e is more than one. Okay? The foci are plus or minus a times e, comma zero; that is the plus-minus sign. The directrices are x equals plus or minus a over e. Focus, a times e; directrix, a over e, same as the ellipse. Right? The transverse axis, along x, has length two a, and the conjugate axis has length two b. Got it? And the latus rectum is two b squared over a, just as for the ellipse.

2:45Focal distances, parametric point, asymptotes

Clear? Now take a point P on the right branch, and let its x coordinate be x. Then S P is e times x, minus a, and S prime P is e times x, plus a, so there are no square roots at all. Okay? Subtract them, and the e times x terms cancel: S prime P minus S P equals two a, every time. On the left branch the two swap roles, which is why the definition has the modulus. With me? Next, the parametric point: a times sec theta, comma, b times tan theta; theta is the Greek letter. It works because sec squared theta, minus tan squared theta, is one. Right? And the asymptotes: y equals plus or minus b over a, times x. These are the grey dashed lines through the centre; the branches keep getting closer to them. Clear? Now a quick one for you. If a equals b, what are the asymptotes, and how do they meet?

3:48Rectangular and conjugate hyperbolas

Yes, exactly! The slopes become one and minus one, so the asymptotes are y equals plus or minus x. One times minus one is minus one, so they are perpendicular. And e squared is one plus one, so e is root two; that is the square root sign. This is the rectangular hyperbola. Okay? The course also writes a rectangular hyperbola as x y equals c squared, with points c times t, comma, c over t. You can check it: c times t, times c over t, gives c squared. Got it? Now swap the signs: y squared over b squared, minus x squared over a squared, equals one. This is the conjugate hyperbola, and its eccentricity is written e prime. Clear? The two eccentricities are linked: one over e squared, plus one over e prime squared, equals one. Keep this handy. Right?

4:43Example 1: e, foci and asymptotes of x2/16 - y2/9 = 1

Okay, Example one, from the course. Find e, the foci, the directrices, the latus rectum and the asymptotes of x squared over sixteen, minus y squared over nine, equals one. Pause the video and try it first. Done? Okay, let's check. Here a is four and b is three. So e squared is one plus nine over sixteen, which is twenty five over sixteen, and e is five over four. Okay? The foci: four times five over four is five, so plus or minus five, comma zero. Right? The directrices: four, divided by five over four, gives sixteen over five. So x equals plus or minus sixteen over five. Clear? The latus rectum: two b squared over a is eighteen over four, which is nine over two. Last, the asymptotes. Four and three, so y equals plus or minus four over three, times x, right? No, no, wait, that slope is upside down! The slope is b over a, three over four. So the asymptotes are y equals plus or minus three over four, times x. All five answers done!

6:03Example 2: centre and foci by completing the squares

Now a harder one from the course. Find the centre and the foci of nine x squared, minus sixteen y squared, minus eighteen x, minus sixty four y, minus one hundred and ninety nine, equals zero. Hint: complete the squares. Pause the video and try it first. Done? Okay, let's check. Group the terms: nine times x squared minus two x, minus sixteen times y squared plus four y, equals one hundred and ninety nine. Let me ask you this: when you complete the square, does the y bracket leave minus sixty four, or plus sixty four? It leaves plus sixty four, so watch that sign! y squared plus four y is y plus two, the whole squared, minus four, and minus sixteen times minus four is plus sixty four. The x bracket leaves minus nine. Now move both to the right side, and their signs flip: one hundred and ninety nine plus nine, minus sixty four, gives one hundred and forty four. Okay? Divide by one hundred and forty four: x minus one, the whole squared, over sixteen, minus y plus two, the whole squared, over nine, equals one. So the centre is one comma minus two. With me? Here a is four and b is three, just like Example one, so e is five over four and a times e is five. So the foci are plus or minus five, comma zero, right? No, no, careful, that is only for a centre at the origin! Here the centre is one comma minus two, so the foci are one plus or minus five, comma minus two. That gives six comma minus two, and minus four comma minus two.

7:57Quick revision: e squared = 1 + b2/a2, foci, asymptotes

So, quick revision. The hyperbola: the modulus of S prime P minus S P is two a, or P S equals e times P M, with e more than one. In standard form, e squared is one plus b squared over a squared, the foci are plus or minus a times e, comma zero, and the directrices x equals plus or minus a over e. The asymptotes are y equals plus or minus b over a, times x, and when the centre moves, the foci move with it. Now go and try the practice problems on this page. All the best!