Tangents, Normals and Chords

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. Take the ellipse x squared over nine plus y squared over four equals one, and a line of slope one. Slide it across, keeping the slope fixed. In how many positions does it just touch the ellipse? Is it one position, two, or none at all? Keep your guess ready. Tangents of the ellipse come up often in JEE papers, so let's begin.

0:27Tangent at a point

Here is our ellipse: x squared over a squared, plus y squared over b squared, equals one, with a greater than b greater than zero; the small two means squared. The tangent at x one, y one, where the small ones are subscripts, is x times x one over a squared, plus y times y one over b squared, equals one. So x squared becomes x times x one. Okay? Here is one from the course. What is the tangent to x squared over sixteen plus y squared over nine equals one, at two comma three root three over two? The tick-shaped sign is the square root. Here x one is two, and y one is three root three over two, so two x over sixteen, plus three root three times y over eighteen, equals one. Multiply by twenty four, and it becomes three x plus four root three times y, equals twenty four. Clear? With the eccentric angle theta, the Greek letter here, the point is a cos theta, comma, b sin theta. Then the tangent is x cos theta over a, plus y sin theta over b, equals one. With me?

1:34Tangents of slope m

Now, if only the slope m is given, the tangents are y equals m x, plus or minus the square root of, a squared times m squared, plus b squared. See the plus or minus sign? Every slope gives two parallel tangents. Okay? Quick question, since students often mix this up. Under the root, which one is multiplied by m squared: a squared, or b squared? Yes, exactly, a squared, the one under x squared, because m goes with x. So y equals m x plus c touches the ellipse when c squared equals a squared times m squared, plus b squared. Got it? The point of contact is minus a squared times m over c, comma, b squared over c, in the big round brackets. Watch that minus sign, okay?

2:21Normals

Next, the normal, perpendicular to the tangent at the point of contact. At x one, y one, it is a squared times x over x one, minus b squared times y over y one, equals a squared minus b squared. The crossed equals sign says x one times y one is not zero, since they sit in the denominators. Clear? Try this one from the course: the normal to x squared over twenty five plus y squared over sixteen equals one, at three comma sixteen over five. Here a squared minus b squared is nine, so twenty five x over three, minus sixteen y over sixteen fifths, equals nine. Multiply by three: twenty five x minus fifteen y, equals twenty seven. Careful, the tangent has a plus in the middle; the normal has a minus. Right? With the eccentric angle, the normal is a times x sec theta, minus b times y cosec theta, equals a squared minus b squared; the c s c on screen is cosec. With me?

3:26The director circle

Now a beautiful result. When two tangents to the ellipse are perpendicular, they always meet on one circle: x squared plus y squared equals a squared plus b squared, called the director circle. Okay? Quick one from the course: what is the director circle of x squared over nine plus y squared over seven equals one? Just add: nine plus seven gives sixteen. So it is x squared plus y squared equals sixteen, a circle of radius four. That was really simple, right?

3:57Tangents and the foci

Now the foci, S and S prime; the little dash is read prime. Drop perpendiculars from both foci to any tangent. Their feet lie on x squared plus y squared equals a squared, the auxiliary circle. Okay? And the product of these two perpendiculars is always b squared, whichever tangent you take, which is a lovely result, isn't it? Next, the reflection property: at any point P, the tangent and the normal bisect the angles between S P and S prime P, okay? And the greatest distance of any normal from the centre is a minus b. Right?

4:32Chords: T = 0 and T = S1

Two short names now, as for the circle. Capital S, with the three-line sign for is defined as, is x squared over a squared plus y squared over b squared, minus one; S one is its value at x one, y one. Capital T is x times x one over a squared, plus y times y one over b squared, minus one. Okay? From a point x one, y one outside the ellipse, draw the two tangents. The chord joining their points of contact, the chord of contact, is T equals zero. Right? And the chord with mid-point x one, y one is T equals S one. Remember: chord of contact, T equals zero; mid-point given, T equals S one. Clear?

5:15Example 1: tangents of slope 1 to x2/9 + y2/4 = 1

Now, remember our opening puzzle? Here it is as Example one from the course. Find the tangents of slope one to x squared over nine plus y squared over four equals one. Pause the video and try it first. Done? Okay, let's check. Here a squared is nine, b squared is four, and m is one. So c squared is nine plus four, which gives thirteen. With me? So c is root thirteen, and the tangent is y equals x plus root thirteen. That is the answer, isn't it? No, no, wait, there is a catch here: since c squared is thirteen, c is plus or minus root thirteen. So there are two tangents, y equals x plus root thirteen, and y equals x minus root thirteen. That answers our puzzle: two positions. Forget the minus sign, and you lose half the answer. Got it?

6:08Example 2: chord of x2/16 + y2/9 = 1 bisected at (2, 1)

Now a harder one, Example two from the course. Find the chord of x squared over sixteen plus y squared over nine equals one, bisected at two comma one. Here is a hint: two comma one is its mid-point. Pause the video and try it first. Done? Okay, let's check. Many students write the tangent-style line straight away: two x over sixteen plus y over nine equals one. Is that right? Oh no, careful here, that is T equals zero, the chord of contact formula, not the mid-point formula. Multiply by seventy two, and it becomes nine x plus eight times y, equals seventy two. Put in two comma one: eighteen plus eight gives twenty six, not seventy two. It misses its own mid-point! Right? The mid-point is given, so we need T equals S one, at two comma one. Two x over sixteen, plus y over nine, minus one, equals four over sixteen, plus one over nine, minus one. The minus ones cancel. With me? So x over eight plus y over nine equals one quarter plus one ninth, which is thirteen over thirty six. Clear? Multiply by seventy two: nine x plus eight times y, equals twenty six. Quick check: nine times two plus eight times one gives twenty six, so it passes through two comma one, and that's our chord!

7:31Quick revision: tangent, normal, director circle, chords

So, quick revision. The tangent at x one, y one is x times x one over a squared, plus y times y one over b squared, equals one, and y equals m x plus c touches when c squared equals a squared times m squared, plus b squared. The normal has a minus in the middle; perpendicular tangents meet on the director circle. Chord of contact, T equals zero; given mid-point, T equals S one. Now try the practice problems on this page. All the best!