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Chapter 15: Problem 1426

The curve represented by \(\mathrm{x}=3(\cos \mathrm{t}+\sin \mathrm{t})\); \(\mathrm{y}=4(\cos \mathrm{t}-\sin \mathrm{t})\) is (a) circle (b) parabola (c) ellipse (d) hyperbola

Short Answer

Expert verified
The curve represented by the given parametric equations is an ellipse, because after eliminating the parameter \(t\) and simplifying the expression, we obtain an equation of the form \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) with \(a^2 = 1\) and \(b^2 = \frac{9}{2}\). Therefore, the correct answer is (c) ellipse.

Step by step solution

01

Express \(\cos t\) and \(\sin t\) in terms of \(x\) and \(y\)

From the given parametric equations, we can express \(\cos t\) and \(\sin t\) in terms of \(x\) and \(y\): \[\cos t = \frac{x}{3} - \frac{y}{4}\] \[\sin t = \frac{x}{3} + \frac{y}{4}\]
02

Square and sum the two equations

Now, we square both equations and sum them to eliminate the parameter \(t\): \[\left(\cos t\right)^2 + \left(\sin t\right)^2 = \left(\frac{x}{3} - \frac{y}{4}\right)^2 + \left(\frac{x}{3} + \frac{y}{4}\right)^2\] Since \(\cos^2 t + \sin^2 t = 1\), we can simplify the above equation to obtain the type of curve represented by these parametric equations.
03

Simplify the resulting equation and determine the type of curve

\begin{align*} 1 &= \left(\frac{x}{3} - \frac{y}{4}\right)^2 + \left(\frac{x}{3} + \frac{y}{4}\right)^2 \\ &= \frac{x^2}{9} - \frac{xy}{6} + \frac{y^2}{16} + \frac{x^2}{9} + \frac{xy}{6} + \frac{y^2}{16}\\ &= \frac{2x^2}{9} + \frac{2y^2}{16} \end{align*} Divide both sides by 2, and then multiply by 9 and 16 to eliminate the fractions: \[9\left(\frac{x^2}{9}\right) + 16\left(\frac{y^2}{16}\right) = 9(1)\] \[\frac{x^2}{1} + \frac{y^2}{\frac{9}{2}} = 1\] This is an equation of an ellipse because it is in the form \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), with \(a^2 = 1\) and \(b^2 = \frac{9}{2}\). So, the correct answer is: (c) ellipse

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Most popular questions from this chapter

If \(\mathrm{P}(\mathrm{m}, \mathrm{n})\) is a point on an ellipse \(\left(\mathrm{x}^{2} / \mathrm{a}^{2}\right)+\left(\mathrm{y}^{2} / \mathrm{b}^{2}\right)=1\) with foci \(\mathrm{S}\) and \(\mathrm{S}^{\prime}\) and eccentricity e, then area of \(\mathrm{SPS}^{\prime}\) is \(\ldots \ldots \ldots\) (a) \(\mathrm{ae} \sqrt{\left(a^{2}-\mathrm{m}^{2}\right)}\) (b) \(\mathrm{ae} \sqrt{\left(b^{2}-\mathrm{m}^{2}\right)}\) (c) \(b e \sqrt{\left(b^{2}-m^{2}\right)}\) (d) be \(\sqrt{\left(a^{2}-m^{2}\right)}\)

If \(\mathrm{AB}\) is a double ordinates of the hyperbola \(\left(\mathrm{x}^{2} / \mathrm{a}^{2}\right)-\left(\mathrm{y}^{2} / \mathrm{b}^{2}\right)=1\) such that \(\mathrm{OAB}\) is an equilateral triangle, \(O\) being the centre of the hyperbola, then the eccentricity e of the hyperbola satisfies. (a) \(1<\mathrm{e}<(2 / \sqrt{3})\) (b) \(\mathrm{e}<(1 / \sqrt{3})\) (c) \(\mathrm{e}>(\sqrt{3} / 2)\) (d) \(e>(2 / \sqrt{3})\)

The triangle \(P Q R\) is inscribed in the circle \(x^{2}+y^{2}=25 .\) If \(Q\) and \(R\) have coordinates \((3,4)\) and \((-4,3)\) respectively, then \(\angle Q P R\) is equal to (a) \((\pi / 2)\) (b) \((\pi / 3)\) (c) \((\pi / 4)\) (d) \((\pi / 6)\)

The equation of the common tangent to the curves \(\mathrm{y}^{2}=8 \mathrm{x}\) and \(\mathrm{xy}=-1\) is (a) \(9 x-3 y+2=0\) (b) \(2 \mathrm{x}-\mathrm{y}+1=0\) (c) \(x-2 y+8=0\) (d) \(x-y+2=0\)

If the line \(\mathrm{y}=1-\mathrm{x}\) touches the curve \(\mathrm{y}^{2}-\mathrm{y}+\mathrm{x}=0\), then the point of contact is (a) \((0,1)\) (b) \((1,0)\) (c) \((1,1)\) (d) \([(1 / 2),(1 / 2)]\)
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