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Chapter 10: Problem 35

Find the equation of the tangent line to the given curve at the given point. $$ \frac{x^{2}}{27}+\frac{y^{2}}{9}=1 \text { at }(3, \sqrt{6}) $$

Short Answer

Expert verified
The equation of the tangent line is \( y = -\frac{1}{\sqrt{6}} x + \frac{9}{\sqrt{6}} \).

Step by step solution

01

Identify the Given Curve

The given curve is an ellipse represented by the equation \(\frac{x^2}{27} + \frac{y^2}{9} = 1\). This equation represents an ellipse centered at the origin.
02

Differentiate Implicitly

To find the slope of the tangent line, differentiate the equation \(\frac{x^2}{27} + \frac{y^2}{9} = 1\) with respect to \(x\). This yields \(\frac{2x}{27} + \frac{2y}{9} \cdot \frac{dy}{dx} = 0\).
03

Solve for \(\frac{dy}{dx}\)

Rearrange the differentiated equation to solve for \(\frac{dy}{dx}\):\[ \frac{dy}{dx} = -\frac{9x}{27y} = -\frac{x}{3y} \]
04

Substitute the Given Point

Substitute the point \((3, \sqrt{6})\) into the equation for \(\frac{dy}{dx}\):\[ \frac{dy}{dx} = -\frac{3}{3\sqrt{6}} = -\frac{1}{\sqrt{6}} \]This is the slope of the tangent line at the point \((3, \sqrt{6})\).
05

Use the Point-Slope Form

With the slope \(-\frac{1}{\sqrt{6}}\) and the point \((3, \sqrt{6})\), use the point-slope form of a line: \[ y - y_1 = m(x - x_1) \]Substitute \( (x_1, y_1) = (3, \sqrt{6}) \) and \( m = -\frac{1}{\sqrt{6}} \):\[ y - \sqrt{6} = -\frac{1}{\sqrt{6}}(x - 3) \]
06

Simplify the Equation

Expand and simplify the equation derived in Step 5:\[ y - \sqrt{6} = -\frac{1}{\sqrt{6}}x + \frac{3}{\sqrt{6}} \]\[ y = -\frac{1}{\sqrt{6}}x + \frac{3}{\sqrt{6}} + \sqrt{6} \]We find a common denominator for the constant terms:\[ y = -\frac{1}{\sqrt{6}}x + \frac{3 \cdot \sqrt{6}}{6} + \frac{6}{\sqrt{6}} \]Combine and simplify:\[ y = -\frac{1}{\sqrt{6}}x + \frac{3 + 6}{\sqrt{6}} \]y = -\frac{1}{\sqrt{6}} x + \frac{9}{\sqrt{6}}.

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Key Concepts

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

Ellipse
An ellipse is a geometric shape that looks like a stretched-out circle. Imagine taking a circle and pressing down on it to elongate it along two axes. It has two main axes: the major axis, which is the longest diameter, and the minor axis, which is the shortest. The equation \(\frac{x^2}{27} + \frac{y^2}{9} = 1\) represents an ellipse centered at the origin, with its major axis along the x-direction and its minor axis along the y-direction.- For the given ellipse, the length of the semi-major axis is \(\sqrt{27}\), while the semi-minor axis is \(\sqrt{9} = 3\). An ellipse can be visualized as a distorted circle, and its unique shape depends on the coefficients in the equation.
Differentiation
Differentiation is a fundamental concept in calculus used to find the rate at which a quantity changes. In the context of an ellipse, differentiation helps us find the slope of the tangent line at any given point on its curve.The process involves finding the derivative of the ellipse's equation with respect to \(x\). This represents how much \(y\) changes for a small change in \(x\). In our example, we start by differentiating \(\frac{x^2}{27} + \frac{y^2}{9} = 1\) to obtain the slope of the tangent line via implicit differentiation.
Point-Slope Form
The point-slope form of a linear equation is used to write the equation of a line when you know both a point on the line and its slope. The equation is given as:\[y - y_1 = m(x - x_1)\]where \((x_1, y_1)\) is the known point and \(m\) is the slope of the line.In our problem, after finding the slope \(-\frac{1}{\sqrt{6}}\) at the point \((3, \sqrt{6})\), we plug these values into the point-slope form to get the equation of the tangent line. It's a quick and efficient method to derive the equation when these parameters are known.
Implicit Differentiation
Implicit differentiation is a technique used in calculus when dealing with equations where \(y\) cannot be easily solved for \(x\). In such cases, instead of solving the equation explicitly for \(y\) and then differentiating, we differentiate both sides of the equation with respect to \(x\), treating \(y\) as an implicit function of \(x\).In our example, we differentiate the ellipse equation directly:\[\frac{2x}{27} + \frac{2y}{9} \cdot \frac{dy}{dx} = 0\]This allows us to solve for \(\frac{dy}{dx}\), the derivative of \(y\) with respect to \(x\), which is crucial for finding the slope of the tangent line. Implicit differentiation simplifies the process of finding derivatives for complex or implicit relationships.

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Most popular questions from this chapter

Assume that a planet of mass \(m\) is revolving around the sun (located at the pole) with constant angular momentum \(m r^{2} d \theta / d t\). Deduce Kepler's Second Law: The line from the sun to the planet sweeps out equal areas in equal times.

The position of a comet with a highly eccentric elliptical orbit \((e\) very near 1\()\) is measured with respect to a fixed polar axis (sun is at a focus but the polar axis is not an axis of the ellipse) at two times, giving the two points \((4, \pi / 2)\) and \((3, \pi / 4)\) of the orbit. Here distances are measured in astronomical units \((1 \mathrm{AU} \approx 93\) million miles). For the part of the orbit near the sun, assume that \(e=1\), so the orbit is given by $$r=\frac{d}{1+\cos \left(\theta-\theta_{0}\right)}$$ (a) The two points give two conditions for \(d\) and \(\theta_{0}\). Use them to show that \(4.24 \cos \theta_{0}-3.76 \sin \theta_{0}-2=0\) (b) Solve for \(\theta_{0}\) using Newton's Method. (c) How close does the comet get to the sun?

Find all points on the cardioid \(r=a(1+\cos \theta)\) where the tangent line is (a) horizontal, and (b) vertical.

Let \(r_{1}\) and \(r_{2}\) be the minimum and maximum distances (perihelion and aphelion, respectively) of the ellipse \(r=\) \(e d /\left[1+e \cos \left(\theta-\theta_{0}\right)\right]\) from a focus. Show that (a) \(r_{1}=e d /(1+e), r_{2}=e d /(1-e)\), (b) major diameter \(=2 e d /\left(1-e^{2}\right)\) and minor diameter \(=\) \(2 e d / \sqrt{1-e^{2}}\)

Name the curve with the given polar equation. If it is a conic, give its eccentricity. Sketch the graph. $$ r=\frac{6}{4-\cos \theta} $$
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