Problem 77
Question
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 $$ \frac{x x_{0}}{a^{2}}+\frac{y y_{0}}{b^{2}}=1 $$
Step-by-Step Solution
Verified Answer
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 \((x_0, y_0)\).
1Step 1: Find the general equation of the ellipse
The given ellipse has the general equation:
$$
\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1
$$
2Step 2: 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:
$$
m = -\frac{2x_0}{2y_0}\frac{b^2}{a^2}
$$
3Step 3: 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:
$$
\frac{x_0 x}{a^2} + \frac{y_0 y}{b^2} = 1
$$
4Step 4: 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
$$
\frac{x x_0}{a^2} + \frac{y y_0}{b^2} = 1
$$
Key Concepts
Implicit DifferentiationSlope of EllipsePoint-Slope Form of a LineEquation of an Ellipse
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.
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.
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.
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
Other exercises in this chapter
Problem 75
Show that the polar equation $$r^{2}-2 r r_{0} \cos \left(\theta-\theta_{0}\right)=R^{2}-r_{0}^{2}$$ describes a circle of radius \(R\) whose center has polar c
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Find parametric equations (not unique) of the following ellipses (see Exercises \(75-76\) ). Graph the ellipse and find a description in terms of \(x\) and \(y.
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Find an equation of the line tangent to the hyperbola \(x^{2} / a^{2}-y^{2} / b^{2}=1\) at the point \(\left(x_{0}, y_{0}\right)\)
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