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anonymous
 5 years ago
Determine the point on the hyperbola −3x2+2y2=10 closest to the point (4, 0).
anonymous
 5 years ago
Determine the point on the hyperbola −3x2+2y2=10 closest to the point (4, 0).

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You need to form the distance function between an arbitrary point on the hyperbola and the point, (4,0), and then determine the minimum using calculus. \[D^2=(x4)^2+(y0)^2\]Using implicit differentiation, you get,\[2D \frac{dD}{dx}=2(x4)+2y \frac{dy}{dx} \rightarrow \frac{dD}{dx}=\frac{x4+yy'}{D}\]Since you're finding extrema, you need to find x such that dD/dx = 0. Hence,\[\frac{dD}{dx}=\frac{x4+yy'}{D}=0 \rightarrow x4+yy'=0\]Now, before going crazy with solving for y and y' in your original equation, I suggest we take the derivative of the equation to see if we can get any relationship between yy' and x (to cut down work). So, \[\frac{d}{dx}3x^2+2y^2=\frac{d}{dx}10 \rightarrow 6x+4yy'=0\]so that\[yy'=\frac{6x}{4}=\frac{3x}{2}\]Substitute this relationship into the last expression we had for dD/dx, to get,\[x4+\frac{3x}{2}=0 \rightarrow x=\frac{8}{5}\]When you substitute this xvalue into your equation for the hyperbola and solve for y, you get,\[y=\pm \frac{\sqrt{221}}{5}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You'll have two points on the hyperbola that are minimal in distance to the point (4,0); that is,\[(\frac{8}{5}, \frac{\sqrt{221}}{5})\]and\[(\frac{8}{5}, \frac{\sqrt{221}}{5})\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If you'd like to thank me, please click the 'Become a fan' link next to my name (this took a long time to write out) :p

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0haha, I seriously appreciate it! :) I became a fan.
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