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Chapter 10: Problem 77

Show that an equation of the line tangent to the ellipse \(x^{2} / a^{2}+y^{2} / b^{2}=1\) at the point \(\left(x_{0}, y_{0}\right)\) is $$ \frac{x x_{0}}{a^{2}}+\frac{y y_{0}}{b^{2}}=1 $$

Short Answer

Expert verified
Question: Show that the given equation is tangent to the ellipse at point \((x_0, y_0)\). Given Equation: \(\frac{x x_0}{a^2} + \frac{y y_0}{b^2} = 1\) Ellipse Equation: \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) Answer: We have shown that the derived tangent line equation, \(\frac{x_0 x}{a^2} + \frac{y_0 y}{b^2} = 1\), matches the given equation. Therefore, the given equation is tangent to the ellipse at point \((x_0, y_0)\).

Step by step solution

01

Find the general equation of the ellipse

The given ellipse has the general equation: $$ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $$
02

Determine the slope of ellipse at \((x_0, y_0)\)

Now we need to find the slope of the ellipse at point \((x_0, y_0)\). To do this, we will first implicitly differentiate the ellipse equation with respect to \(x\): $$ \frac{2x}{a^2} + \frac{2y}{b^2}\frac{dy}{dx} = 0 $$ Now solve for \(\frac{dy}{dx}\) to find the slope: $$ \frac{dy}{dx} = -\frac{2x}{2y}\frac{b^2}{a^2} $$ Now, we can plug in \((x_0, y_0)\) to get the slope at that specific point: $$ m = -\frac{2x_0}{2y_0}\frac{b^2}{a^2} $$
03

Obtain the equation of the tangent line to ellipse

Using the point-slope form, we can write the tangent line equation at \((x_0, y_0)\). The point-slope form is \(y - y_1 = m(x - x_1)\). $$ y - y_0 = m(x - x_0) $$ Where \(m\) is the slope we calculated in Step 2. Plug \(m\) into the equation and simplify: $$ y - y_0 = -\frac{2x_0}{2y_0}\frac{b^2}{a^2}(x - x_0) $$ Rewrite the equation and simplify a bit more: $$ \frac{b^2}{a^2}\left(\frac{x_0(x - x_0)}{y_0}\right) + y = y_0 $$ Rearrange the terms: $$ \frac{x_0 x}{a^2} + \frac{y_0 y}{b^2} = 1 $$
04

Show that our derived tangent equation matches the given equation

The tangent line equation we derived is: $$ \frac{x_0 x}{a^2} + \frac{y_0 y}{b^2} = 1 $$ Which matches the given equation, thus proving that the equation of the line tangent to the ellipse at the point \((x_0, y_0)\) is $$ \frac{x x_0}{a^2} + \frac{y y_0}{b^2} = 1 $$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Implicit Differentiation
Implicit differentiation is a technique used in calculus to find the derivative of a function without explicitly solving it for one variable. When dealing with equations like the ellipse equation \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), both x and y are interrelated, making it impossible to isolate y and differentiate directly.

The process begins by differentiating both sides of the equation with respect to x, treating y as an implicit function of x. For instance, when we differentiate \(y^2\), we apply the chain rule: \(2y \frac{dy}{dx}\). The derivative \(\frac{dy}{dx}\) represents the slope of the tangent line at a given point on the ellipse, and through this method, we can compute it without the explicit expression for y.
Slope of Ellipse
The slope of an ellipse at any point is essential for understanding the direction in which the ellipse is moving at that point. It's particularly useful when you want to find the equation of a tangent line at a specific point \(\left(x_0, y_0\right)\) on the ellipse.

By using implicit differentiation, the slope \(m\) at a point on the ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) is given by \(m = -\frac{x_0}{y_0}\frac{b^2}{a^2}\). This negative reciprocal relationship between \(x_0\) and \(y_0\) reflects how the tangent line is perpendicular to the radius of the ellipse at that point, which makes sense since tangents are always perpendicular to radii at the point of tangency.
Point-Slope Form of a Line
The point-slope form of a line is a straightforward way to write the equation of a line when you have a point and the slope. This form is written as \(y - y_1 = m(x - x_1)\), where \(m\) is the slope and \(\left(x_1, y_1\right)\) is a point on the line.

For tangent lines to an ellipse, once we find the slope \(m\) at a particular point \(\left(x_0, y_0\right)\), we can substitute these values into the point-slope form. Then, algebraic manipulations allow us to rearrange the equation into a more familiar linear equation format. It's a useful formula because it directly uses the properties of the point and slope without requiring any additional transformations.
Equation of an Ellipse
The equation of an ellipse represents the set of all points \(\left(x, y\right)\) that form an elliptical shape when plotted on a graph. The standard form of an ellipse equation is \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), where

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Most popular questions from this chapter

Graph the following conic sections, labeling the vertices, foci, directrices, and asymptotes (if they exist). Use a graphing utility to check your work. $$r=\frac{1}{2-\cos \theta}$$

An epitrochoid is the path of a point on a circle of radius \(b\) as it rolls on the outside of a circle of radius \(a\). It is described by the equations $$\begin{array}{l}x=(a+b) \cos t-c \cos \left(\frac{(a+b) t}{b}\right) \\\y=(a+b) \sin t-c \sin \left(\frac{(a+b) t}{b}\right)\end{array}$$ Use a graphing utility to explore the dependence of the curve on the parameters \(a, b,\) and \(c.\)

A focal chord of a conic section is a line through a focus joining two points of the curve. The latus rectum is the focal chord perpendicular to the major axis of the conic. Prove the following properties. Let \(L\) be the latus rectum of the parabola \(y^{2}=4 p x,\) for \(p>0\) Let \(F\) be the focus of the parabola, \(P\) be any point on the parabola to the left of \(L,\) and \(D\) be the (shortest) distance between \(P\) and \(L\) Show that for all \(P, D+|F P|\) is a constant. Find the constant.

A focal chord of a conic section is a line through a focus joining two points of the curve. The latus rectum is the focal chord perpendicular to the major axis of the conic. Prove the following properties. The length of the latus rectum of an ellipse centered at the origin is \(2 b^{2} / a=2 b \sqrt{1-e^{2}}\)

The butterfly curve of Example 8 is enhanced by adding a term: $$r=e^{\sin \theta}-2 \cos 4 \theta+\sin ^{5}(\theta / 12), \quad \text { for } 0 \leq \theta \leq 24 \pi$$ a. Graph the curve. b. Explain why the new term produces the observed effect. (Source: S. Wagon and E. Packel, Animating Calculus, Freeman, 1994)
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