Given the hyperbola $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1,$$ show that the point closest to focus $F(c, 0)$ where $c^2=a^2+b^2$ is the vertex $V(a, 0)$
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Shouldn't $F$ and $V$ be $(c,0)$ and $(a,0)$, respectively? In any case, here is a hint: you can express the distance between a point on the hyperbola and $F$ as a function of $y$. – zuggg Jul 16 '13 at 06:01
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You're right. Fixed it. – hondaman Jul 16 '13 at 06:15
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Solving for $x$ and considering the right branch where $x>0$ $$x = a\sqrt{1+\frac{y^2}{b^2}}$$ The distance between $F(c,0)$ and $P\left(a\sqrt{1+\frac{y^2}{b^2}}\;,\; y\right)$ is $$\sqrt{\left(\frac{a}{b}\sqrt{b^2+y^2} - c\right)^2 + y^2}$$ The derivative of the expression under the radical is $$2y\left(\frac{a^2}{b^2} + \frac{ac}{b\sqrt{b^2+y^2}} + 1\right)$$
The derivative has only one solution, $y=0$ and by the derivative test, we can see that the minimum value for the distance occurs at $y=0$, $x=a$.
hondaman
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Let $F_\pm:=(\pm c,0)$ be the two foci. Then the right half $\gamma_+$ of your hyperbola is the set of all points $P\in{\mathbb R}^2$ for which $$d(P,F_+)=d(P,F_-)-2a\ .$$ It follows that for $P\in \gamma_+$ the quantity $d(P,F_+)$ is smallest when $d(P,F_-)$ is smallest, and the latter is obviously the case at the apex $(a,0)$.
Christian Blatter
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