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anonymous

  • 5 years ago

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

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  1. anonymous
    • 5 years ago
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    You 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. \[D^2=(x-4)^2+(y-0)^2\]Using implicit differentiation, you get,\[2D \frac{dD}{dx}=2(x-4)+2y \frac{dy}{dx} \rightarrow \frac{dD}{dx}=\frac{x-4+yy'}{D}\]Since you're finding extrema, you need to find x such that dD/dx = 0. Hence,\[\frac{dD}{dx}=\frac{x-4+yy'}{D}=0 \rightarrow x-4+yy'=0\]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,\[x-4+\frac{3x}{2}=0 \rightarrow x=\frac{8}{5}\]When you substitute this x-value into your equation for the hyperbola and solve for y, you get,\[y=\pm \frac{\sqrt{221}}{5}\]

  2. anonymous
    • 5 years ago
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    You'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\[(\frac{8}{5}, -\frac{\sqrt{221}}{5})\]

  3. anonymous
    • 5 years ago
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    If 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

  4. anonymous
    • 5 years ago
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    haha, I seriously appreciate it! :) I became a fan.

  5. anonymous
    • 5 years ago
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    np

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