Mathematical Immaturity

14.21 Miscellaneous Review Exercises

• Let $V$ be one of the two vertices of a hyperbola whose transverse axis has length $2a$ and whose eccentricity is 2. Let $P$ be a point on the same branch as $V.$ Denote by $A$ the area of the region bounded by the hyperbola and the line segment $VP,$ and let $r$ be the length of $VP.$ (a) Place the coordinate axes in a convenient position and write an equation for the hyperbola. (b) Express the area $A$ as an integral and, without attempting to evaluate this integral, show that $Ar^{-3}$ tends to a limit as the point $P$ tends to $V.$ Find this limit.
• Recall from Section 13.22 that for a conic of eccentricity $e$ and distance from the focus to directrix $d,$ we have $a = ed/(1 - e^2).$ Then, if $N$ is the unit vector normal to the directrix, the vertices of the conic are located at $\pm aN.$ For a hyperbola, the transverse axis is the line segment joining these two points.
• (a) Let the $x$- and $y$-axes be the coordinate axes, with $N = \mathbf{i}$ being the vector normal to the directrix. The standard form equation for a conic with $e \neq 1$ is then: \begin{align*} \frac{x^2}{a^2} + \frac{y^2}{a^2(1-e^2)} &= 1 \end{align*} If $e = 2,$ then $3x^2 - y^2 = 3a^2 \quad \blacksquare$ (b) From our choice of coordinate axes and normal vector, we know from Section 13.22 that the vertices of the hyperbola are located at $(-a, 0)$ and $(a, 0).$ Let $V = (a, 0)$ and let $P = (x_0, y_0)$ be a point on the right branch in the first quadrant. Then, the line segment connecting $V$ to $P$ can be parameterized by $x$ as \begin{align*} y &= \frac{y_0}{x_0 - a}(x - a) \end{align*} Then, the integral $A$ is the area under the hyperbola in quadrant 1 minus the area under the line $y$ from $x = a$ to $x = x_0$ \begin{align*} A &= \int_a^{x_0} \sqrt{3(x^2 - a^2)} - \frac{y_0}{x_0 - a}(x - a)\,dx \end{align*} The length $r$ of the line segment $VP$ is the norm $\|P - V\| = \sqrt{(x_0 - a)^2 + y_0^2}.$ But since $y_0$ is on the hyperbola, we can write $r = \sqrt{(x_0 - a)^2 + 3(x_0^2 - a^2)}.$ Now, let $h = x_0 - a,$ $h > 0.$ We can express $A$ as: \begin{align*} A &= \int_a^{a + h} \sqrt{3(x^2 - a^2)} - \frac{\sqrt{3h^2 + 6ha}}{h}(x - a)\,dx \\ \\ &= \int_a^{a + h} \sqrt{3(x^2 - a^2)}\,dx - \frac{h}{2}\sqrt{3h^2 + 6ha} \end{align*} Making the substitution $u = x - a,$ $u < h,$ \begin{align*} A &= \int_0^h \sqrt{3(u^2 + 2au)}\,du - \frac{h}{2}\sqrt{3h^2 + 6ha} \end{align*} And with $h = x_0 - a,$ $r$ becomes $\sqrt{4h^2 + 6ha},$ making $r^3 = \left(4h^2 + 6ha\right)^{3/2}.$ Then, as $h \to 0,$ $u \to 0,$ and the limit of $Ar^{-3}$ becomes: \begin{align*} Ar^{-3} &= r^{-3}\int_0^h \sqrt{6au}\left[1 + o(1)\right]\,du - \frac{\left[1 + o(1)\right]}{4\left(2h + 3a\right)} \\ \\ &= \frac{2\sqrt{6a}}{3}\left(\frac{h}{4h^2 + 6ha}\right)^{3/2} - \frac{\left[1 + o(1)\right]}{4\left(2h + 3a\right)} \\ \\ &= \frac{2\sqrt{6a}}{3}\left(\frac{1}{4h + 6a}\right)^{3/2} - \frac{\left[1 + o(1)\right]}{4\left(2h + 3a\right)} \\ \\ &= \frac{1}{9a} - \frac{1}{12a} \\ \\ &= \frac{1}{36a} \quad \blacksquare \end{align*}