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Chapter 5: Problem 80

\(x=2-3 \sin \theta, \quad y=3+2 \cos \theta\) \(\left(\frac{x-2}{3}\right)^{2}+\left(\frac{y-3}{2}\right)^{2}=1 \Rightarrow\) eqn of ellipse end pts of major axis \(\rightarrow(-1,3) \&(5,3)\)

Short Answer

Expert verified
Question: Verify that the given parametric equations represent an ellipse and find the endpoints of the major axis. Given parametric equations: \(x=2-3\sin\theta\) and \(y=3+2\cos\theta\). Answer: The given parametric equations represent an ellipse with the Cartesian equation \(\left(\frac{2-x}{3}\right)^2 + \left(\frac{y-3}{2}\right)^2 = 1\). The endpoints of the major axis of this ellipse are \((-1,3)\) and \((5,3)\).

Step by step solution

01

Eliminate the parameter 饾渻

To eliminate the parameter \(\theta\), first Express \(\sin\theta\) and \(\cos\theta\) from the given parametric equations: \(\sin\theta=\frac{2-x}{3}\) and \(\cos\theta=\frac{y-3}{2}\). Now, use the trigonometric identity \(\sin^2\theta + \cos^2\theta = 1\) by plugging in the expressions for \(\sin\theta\) and \(\cos\theta\).
02

Obtain the Cartesian equation

Substitute the expressions for \(\sin\theta\) and \(\cos\theta\) into the trigonometric identity: \(\left(\frac{2-x}{3}\right)^2 + \left(\frac{y-3}{2}\right)^2 = 1\). This is the equation we want to analyze and confirm that it represents an ellipse.
03

Identify the equation as representing an ellipse

In the standard form, the equation of an ellipse centered at point \((h,k)\) with the major axis of length \(2a\) and the minor axis of length \(2b\) is given by: \(\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1\) Comparing this standard form with our equation \(\left(\frac{2-x}{3}\right)^2 +\left(\frac{y-3}{2}\right)^2 = 1\), we can see that the equation is indeed representing an ellipse and is already in standard form, with \((h,k)=(2,3)\), \(a=3\), and \(b=2\).
04

Find the endpoints of the major axis

Since our ellipse has a horizontal major axis (because \(a> b\)), the endpoints of the major axis are \((h \pm a, k)\). Using the values we obtained earlier, the endpoints are \((2 \pm 3, 3)\). So, the endpoints of the major axis of the ellipse are \((-1,3)\) and \((5,3)\).

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Most popular questions from this chapter

\(|\sin 2 \mathrm{x}|=|\mathrm{x}|-\mathrm{a}\) \(|\mathbf{x}|-\mathrm{a}>1 \quad\) for no solution \(|x|>1+a\)

\(\mathrm{s}=30\left(1-\mathrm{e}^{\mathrm{ln}}\right)\) \(5=30\left(\mathrm{l}-\mathrm{e}^{k}\right)\) \(\mathrm{e}^{k}=\frac{5}{6}\) \(\mathrm{k}=\ln (5 / 6)\)

\(y=2 x^{2}-x+1\) \(\frac{d y}{d x}=4 x_{1}-1\) \(4 x_{1}-1=3\) \(x_{1}=1\) \(y_{1}=2\)

\(x^{3 / 2}+y^{3 / 2}=2 a^{3 / 2}\) \(\frac{3}{2} x^{1 / 2}+\frac{3}{2} y^{1 / 2} \frac{d y}{d x}=0\) \(\frac{\mathrm{dy}}{\mathrm{dx}}=-\sqrt{\frac{\mathrm{x}}{\mathrm{y}}}=-1\) \(\mathrm{y}=\mathrm{x}\) Putting in eqn \(x^{3 / 2}=a^{3 / 2}\) \(\Rightarrow \mathrm{x}=\mathrm{a}, \mathrm{y}=\mathrm{a}\)

\(x^{4}+y^{4}=a^{4}\) \(x^{3}+y^{3} \frac{d y}{d x}=0\) \(\frac{d y}{d x}=\frac{-x^{3}}{y^{3}}\) eqn of tangent \(\rightarrow\) \(\mathrm{y}-\mathrm{y}_{1}=-\frac{\mathrm{x}_{1}^{3}}{\mathrm{y}_{1}^{3}}\left(\mathrm{x}-\mathrm{x}_{1}\right)\) \(\mathrm{p}=\frac{\mathrm{y}_{1}^{4}}{\mathrm{x}_{1}^{3}}+\mathrm{x}_{1}=\frac{\mathrm{x}_{1}^{4}+\mathrm{y}_{1}^{4}}{\mathrm{x}_{1}^{3}}=\frac{\mathrm{a}^{4}}{\mathrm{x}_{1}^{3}}\) \(\mathrm{q}=\mathrm{y}_{1}+\frac{\mathrm{x}_{1}^{4}}{\mathrm{y}_{1}^{3}}=\frac{\mathrm{a}^{4}}{\mathrm{y}_{1}^{3}}\) Now, \(\mathrm{p}^{-4 / 3}+\mathrm{q}^{-4 / 3}=\mathrm{a}^{-4 / 3}\)
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