Show that the vertex is the point on a branch of a hyperbola that is closest to the focus associated with that branch 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 $V(a, 0)$

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 $y=0$, $x=a$.
