Find the equation of the line that is tangent to the ellipse b^2 x^2+a^2 y^2=a^2 b^2 in the firs

step 1 Here we're given the equation of an ellipse. So let's actually turn this into standard form, whi

Find the equation of the line that is tangent to the ellipse...

Key Concepts

Ellipse Geometry

An ellipse is a conic section characterized by two focal points and a smooth closed curve. Its standard equation relates the squares of the x and y coordinates scaled by constants, reflecting the ellipse's dimensions and axes. Understanding the ellipse’s geometry is essential for determining where and how a line may be tangent to it.

Tangent Line to a Curve

A tangent line touches a curve at exactly one point, having the same slope as the curve does at that point. The concept of tangency is crucial in problems where the line must meet the curve under specific conditions, such as minimizing an area or other optimization constraints.

Intercept Form of a Line

The intercept form expresses a line in terms of its x-intercept and y-intercept. This form is particularly useful when a line creates geometrical figures—like triangles—with the coordinate axes, as it directly relates the line’s equation to the dimensions and area of the resulting triangle.

Optimization Using Calculus

Optimization involves finding the minimum or maximum values of a function under certain constraints. In geometric problems, calculus techniques such as differentiation are used to optimize quantities like areas or perimeters by finding the conditions under which these quantities achieve their extremum values.

Implicit Differentiation

Implicit differentiation is employed when differentiating equations in which the variables are intermingled, such as in the equation of an ellipse. This technique allows for finding the derivative (and hence the slope of the tangent line) even when the function is not explicitly solved for one variable.

Transcript

00:01 Here we're given the equation of an ellipse.

00:04 So let's actually turn this into standard form, which means we're going to divide this entire equation by a squared b squared.

00:11 Okay, so then we end up with x squared over a squared plus y squared over b squared is equal to one.

00:21 So now the question is we want to find the smallest triangle possible formed by the equation of a line that's tangent to this ellipse in the first quadrant and the x and y axis.

00:37 Okay, so let's first draw what that looks like.

00:42 So here we have our x and y coordinates, and then here we have our ellipse.

00:50 So something like this.

00:52 Okay, so here we have our, oh sorry, that's positive, like that.

01:02 So now the tangent line in the first quadrant, let's say to some point would be here.

01:08 And then if we draw our tangent line, it looks something like this.

01:12 So we're interested in this triangular area.

01:18 And we want to minimize that.

01:20 So the first thing that we need to do is actually find the equation of the tangent line.

01:24 And also do notice that we can actually write x and y in terms of t.

01:30 So we can write in parametric form.

01:34 So let's write that on the side.

01:36 So note that x comma y is really acose theta comma b sine theta, right? because if you actually substitute those in, you'll get that this is equal to one.

01:52 So we want to do this because it'll be a little bit easier later when we try to find the derivative.

01:58 So then we don't end up with, you know, bunch of square roots and stuff.

02:02 So let's go ahead and do that.

02:03 So we'll take our equation of the ellipse and then we are going to implicit differentiate both sides.

02:09 Okay, but then we'll keep it so that y is a function of x.

02:13 Okay, so let's go ahead and do our implicit differentiation.

02:17 So this is going to be 2x over 8 squared plus 2y y prime, which is really d y d x over b squared, and this is equal to zero.

02:37 Okay, so now we just want to isolate d, y, d, x.

02:41 Okay, so then we get 2y, d, d, x is equal to negative 2x over a squared, and then times b squared.

02:53 So b squared's on top.

02:55 And then divide both sides by 2y, and we end up with d, y, d, d, x, is equal to negative b, squared x over a squared y okay perfect so now we know what the derivative is with respect to x and y so now we're going to assume that we have a point on the ellipse so let's say we have this point here and that point we're going to call that a cos theta b sine theta okay so the equation of the tangent line with this slow going through this point, we can use our point slope formula.

03:42 So that's going to give us, let's go ahead and make that.

03:48 That's y minus b sine theta is equal to dydx.

03:55 So that's b squared, a squared, y.

04:02 And then we would multiply by y minus our, oh, sorry, actually, that's not x on y, because we want that that would be the derivative right at this point.

04:16 So this would actually be a close theta.

04:22 And then this is over a squared.

04:24 The y value is b sine theta.

04:29 And then this is x minus the x value, which is a cost theta.

04:36 Okay, so let's go ahead and simplify this a little bit.

04:39 So the a will cancel out with this.

04:41 The b will cancel out with that.

04:43 And we get y minus b sine theta is equal to negative b cos theta over a sine theta times x and then plus b cos theta or b co square theta over oh and there's an a on top which we'll cancel out with the a on the bottom so that should be sine theta on the bottom.

05:21 Yes.

05:21 Okay.

05:24 So now we want to simplify this just a little bit more.

05:28 So this is going to be why let's put it over, should we put it over common denominator first? let's see.

05:39 Yeah, let's multiply everything through by sine theta first.

05:43 So then this will be y sine theta minus b sine square theta is equal to negative b cos theta over a.

05:54 Plus b coast square theta.

05:57 There we go.

05:59 And then add b sine square theta onto both sides, and we get y sine theta...