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

1. anonymous

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

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

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

haha, I seriously appreciate it! :) I became a fan.

5. anonymous

np