Mixed Problems: Set 3 (Conics)

Transcript

0:00Introduction

Hello everyone! Okay, a quick challenge before we start. Find the ellipse with foci at zero comma plus or minus four, and eccentricity four fifths. Many students write a times e equals four, so a is five, b squared is twenty five minus sixteen, which is nine, and the answer is x squared over twenty five, plus y squared over nine, equals one. Done in ten seconds, right? Hmm, something is off here. We will test it in Example two. This is Set three of our mixed problems, on conics, and such quick questions come up often in JEE Main. Three tools first, then two problems. Let's begin!

0:41Tool 1: the standard ellipse

So, tool one, the standard ellipse. In the course's figure, the dark curve is the ellipse, S and S prime are its foci, with the little dash read as prime, and the red dashed lines are the directrices. The small a and b in the figure mark the two half axes. The point P with its two orange lines, and the dashed grey circle, are not needed today. The equation is x squared over a squared, plus y squared over b squared, equals one; the small two means squared. The greater than signs say a is bigger than b, and b is bigger than zero. Okay? Next, b squared equals a squared, times, in round brackets, one minus e squared. Here e is the eccentricity, and the two less than signs say it lies between zero and one. Right? The foci are plus or minus a times e, comma zero; that is the plus-minus sign. And the directrices are x equals plus or minus a over e. Focus, a times e; directrix, a over e. Clear? Now let me ask you this: all of this assumed a is bigger than b. What changes if b is bigger, so the bigger denominator is under y squared? Yes, exactly, the roles of a and b swap! The major axis is now along the y axis, so a squared equals b squared, times one minus e squared. And the foci move to zero comma plus or minus b times e. With me? So here is a topper's habit: before any formula, compare the two denominators. The bigger one gives the major axis, and the foci sit on it. Got it?

2:21Tool 2: the standard hyperbola

Now tool two, the hyperbola. In the figure, the two dark curves are its branches, with the foci S and S prime, the red dashed directrices, and again a point P with two orange lines, which we can skip. Its equation has just one change: x squared over a squared, minus y squared over b squared, equals one. And here the eccentricity e is greater than one. Okay? So tell me, if e is bigger than one, how should we write b squared, so that it stays positive? Yes, exactly, flip the bracket! b squared equals a squared, times e squared minus one. Ellipse, one minus e squared; hyperbola, e squared minus one. Right? The foci are still plus or minus a times e, comma zero. And the dashed grey lines in the figure are the asymptotes, y equals plus or minus b over a, times x. Clear?

3:18Tool 3: when does y=mx+c touch the curve?

Now tool three, the most useful one here. We want the condition for the line y equals m x plus c to just touch the curve. Here m is the slope, and c is the intercept on the y axis. For the parabola y squared equals four times a times x, the answer is neat: c equals a over m. Okay? For the standard ellipse, the condition is c squared equals a squared times m squared, plus b squared. With me? Now the hyperbola. Its equation is the ellipse with plus y squared changed to minus y squared. So what do you think its condition is? Yes, exactly, c squared equals a squared times m squared, minus b squared. The only change is the sign of b squared, just like in the equations. Ellipse plus, hyperbola minus, okay? And the course adds one condition for the hyperbola. Tangents of slope m exist only when mod m, with the two vertical bars, is greater than b over a. Then c squared comes out positive. So always check the slope first. Got it?

4:26Example 1: when y = x + c touches x2/4 - y2/3 = 1

Alright, the first one is problem twenty two of Set three, marked easy. For what values of c does the line y equals x plus c touch the hyperbola x squared over four, minus y squared over three, equals one? Start your clock, pause the video, and try it first. Done? Okay, let's check. Compare with tool two: a squared is four, and b squared is three. And the slope m is one, the number in front of x. Right? First, let us check the slope. b over a is root three over two; that is the square root sign. Root three is less than two, so b over a is less than one, which is mod m. So tangents of slope one exist. Clear? Now for c squared, is it four plus three, or four minus three? It is four minus three, because this is a hyperbola! c squared equals a squared times m squared, minus b squared: four times one, minus three, which gives one; the dot means times. If you wrote four plus three, which is seven, you used the ellipse formula. Okay? So c squared is one, and c is plus or minus one. Both lines, y equals x plus one, and y equals x minus one, touch the hyperbola. So don't drop the minus one. Got it?

5:46Example 2: ellipse with foci (0, ±4) and e = 4/5

Now a harder one, problem twenty one of Set three, marked medium, and it is our opening challenge. Find the ellipse with foci zero comma plus or minus four, and eccentricity four fifths. Here is a hint: look at where the foci sit. Pause the video and try it first. Done? Okay, let's check. Our challenge said: a times e is four, so a is five, b squared is nine, and the ellipse is x squared over twenty five, plus y squared over nine, equals one. That looks correct, right? Oh no, careful here, because in that answer the bigger denominator, twenty five, is under x squared. So its foci are plus or minus four, comma zero, on the x axis, not on the y axis. With me? Our foci are zero comma plus or minus four, on the y axis. So the major axis is along y, and b is bigger than a; the double arrow means implies. This is exactly the case from tool one. Clear? Now the foci are zero comma plus or minus b times e. So b times e equals four, that is, b times four fifths equals four, which gives b equals five. Okay? Next, a squared equals b squared, times one minus e squared. Open the bracket, and it is b squared, minus b times e, the whole squared. That is twenty five minus sixteen, which gives nine. Got it? So the ellipse is x squared over nine, plus y squared over twenty five, equals one. And a quick check: b times e is five times four fifths, which is four, so the foci are zero comma plus or minus four. So the challenge is solved: the bigger denominator is under y squared, the axis of the foci. If you caught that trap yourself, very well done!

7:37Quick revision: ellipse vs hyperbola, tangency, major axis

So, quick revision. Ellipse, b squared equals a squared times one minus e squared; hyperbola, b squared equals a squared times e squared minus one. The line y equals m x plus c touches the ellipse when c squared equals a squared times m squared, plus b squared, and the hyperbola with minus b squared. And compare the denominators first; the bigger one gives the major axis. Now try the other problems of Set three, and the practice problem on this page. All the best!