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Solve the Problem y=±12xy = \pm \frac { 1 } { 2 } x

Question 1

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Solve the problem.
-In 1911, Ernest Rutherford discovered the nucleus of the atom. Experiments leading to this discovery involved the scattering of alpha particles by the heavy nuclei in gold foil. When alpha particles are thrust towards the
Gold nuclei, the particles are deflected and follow a hyperbolic path. Suppose the minimum distance that the alpha particles get to the gold nucleus is 6 microns, and that hyperbolic path has asymptotes of y=±12xy = \pm \frac { 1 } { 2 } x . Determine an equation of the path of the particle shown.
 Solve the problem. -In 1911, Ernest Rutherford discovered the nucleus of the atom. Experiments leading to this discovery involved the scattering of alpha particles by the heavy nuclei in gold foil. When alpha particles are thrust towards the Gold nuclei, the particles are deflected and follow a hyperbolic path. Suppose the minimum distance that the alpha particles get to the gold nucleus is 6 microns, and that hyperbolic path has asymptotes of  y = \pm \frac { 1 } { 2 } x . Determine an equation of the path of the particle shown.    A)   \frac { x ^ { 2 } } { 36 } + \frac { y ^ { 2 } } { 9 } = 1  B)   \frac { x ^ { 2 } } { 36 } - \frac { y ^ { 2 } } { 9 } = 1  C)   \frac { x ^ { 2 } } { 6 } - \frac { y ^ { 2 } } { 3 } = 1  D)   \frac { y ^ { 2 } } { 36 } - \frac { x ^ { 2 } } { 9 } = 1


A) x236+y29=1\frac { x ^ { 2 } } { 36 } + \frac { y ^ { 2 } } { 9 } = 1
B) x236y29=1\frac { x ^ { 2 } } { 36 } - \frac { y ^ { 2 } } { 9 } = 1
C) x26y23=1\frac { x ^ { 2 } } { 6 } - \frac { y ^ { 2 } } { 3 } = 1
D) y236x29=1\frac { y ^ { 2 } } { 36 } - \frac { x ^ { 2 } } { 9 } = 1

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