Proof Prove Theorem 10.4 by showing that the tangent line to an ellipse at a point P makes equal

Proof Prove Theorem 10.4 by showing that the tangent line to...

Key Concepts

Reflection Property

The reflection property of ellipses states that a ray emanating from one focus and reflecting off the ellipse will pass through the other focus. This is equivalent to the tangent line at any point on the ellipse making equal angles with the lines joining that point to the foci, and it is a key concept in proving related theorems about the ellipse's geometry.

Ellipse

An ellipse is a type of conic section defined as the set of points whose sum of distances from two fixed points (foci) is constant. This geometric property gives the ellipse its distinctive shape and is central to understanding many of its reflective and tangency properties.

Tangent Line

The tangent line to a curve at a given point is the line that just 'touches' the curve at that point without crossing it locally. For an ellipse, the tangent line at a point is critical because it provides information about the instantaneous direction of the curve and is used to explore reflective properties related to the foci.

Slope Calculation

Calculating the slope of a line is essential in analytic geometry as it quantifies the line's steepness and direction. In the context of an ellipse, finding the slope of the tangent line at a specific point typically involves differentiation, and comparing this slope with that of other lines through the point is key to establishing angular relationships.

Angle Between Lines

The angle between two lines in the plane is determined by the difference in their slopes, and there is a specific formula, tan? = |(m1 - m2) / (1 + m1*m2)|, for calculating this angle. This concept is important when comparing the angles formed by the tangent line and lines connecting the point of tangency to the foci.

Transcript

00:01 Okay, what we want to do is to prove theorem 10 .4, which says let p be a point on an ellipse.

00:08 The tangent line to the ellipse at point p makes equal angles with the line through p and the foci.

00:16 So let's go ahead and kind of give us a picture.

00:22 So we're going to just quickly sketch a ellipse.

00:29 And so let's say, and we're going to make the major axis be the horizontal.

00:41 So here is our ellipse, of course, not drawn to scale.

00:46 And here are our fosite at c, 0, and negative c, 0.

00:56 And we're just going to pick some point p on the ellipse.

01:02 And then we are going to try to draw a tangent line as best we can to this point.

01:10 Okay.

01:11 So there is our tangent line to that point.

01:16 And it is going to be, we're going to go ahead and just make it some x, y values.

01:24 And so then if i draw a line from one of the focci to that point, it makes some angle to that tangent line.

01:36 And let's just call him beta.

01:39 And then if i draw a line from the next fosite to that point, then we have a second angle, and we're just going to call that alpha.

01:50 Okay.

01:51 And so we're trying to show that basically the absolute value of beta equals the absolute value of alpha.

02:01 That those two angles are equal.

02:04 And so let's get started.

02:07 So let's, first thing, let's define our ellipse.

02:12 And the way i have it drawn, if this is a and negative a, negative b and b.

02:18 And so it's centered at zero zero.

02:20 So x squared over a squared plus y squared over b squared equal to one.

02:29 Okay.

02:30 And so since we're talking about tangent lines, we know we're going to have to be taking the derivative.

02:34 So let's take the derivative of our ellipse.

02:42 And we get this is equal to 2x over a squared plus 2y over b squared, y prime equal to zero.

02:54 And then let's go ahead and solve for y prime or dy over dx.

02:59 So y prime is equal to negative x b squared over y a squared.

03:08 Okay.

03:10 And then also something, so let's go ahead and continue with that.

03:17 So at that point x ,0, y zero, so at that point p, then we know that that derivative evaluated at this.

03:32 Point p is actually equal to negative x0 or x not times b squared over y not times a squared and this is going to be defined to be m and it's the slope of tangent line at p.

03:57 Okay now what we want to do is we want to find the slope of the slope of the length line at p.

03:58 Okay.

04:00 Now what we want to do is we want to do is we want to find the slope of the line that goes from one fosi to that point.

04:17 And so we're going to go ahead and find that.

04:19 And so of course that would be y0 minus 0 over x not plus c and we're going to find that to be the slope of the first line.

04:30 Okay.

04:31 And then we're going to find the slope of line going from the second fosai to that and that is going to be y not over x not minus c which is going to be our second slope so the slope of that second line okay now we're also told in the directions so let me go back we're also given or told this equation and it said that the tangent of an angle is equal to a slope minus another slope of the lines that are intersect each other over one plus.

05:37 And i think it said a little bit differently in the problem.

05:43 It was written as, let me see, and it's going to be the absolute value.

05:50 I think it was m1 minus m2, so let's go ahead.

05:53 So, oops, m1 minus m2, and of course it's the absolute value of plus m1, m2...

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