Mathematical Immaturity

13.25. Miscellaneous review exercises on conic sections

• Find a Cartesian equation for a hyperbola passing through the origin, given that its asymptotes are the lines $y = 2x + 1$ and $y = -2x + 3$
• Write the Cartesian equation for the hyperbola assuming its directrices are horizontal lines, ie. $$\frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1$$Rearranging variables, we see that $$y = \pm \frac{a}{\left|b\right|}\sqrt{b^{2} + x^{2}}$$ For large positive $x,$ the value $\sqrt{b^{2} + x^{2}}$ is nearly equal to $x,$ thus the value $y = \pm \frac{a}{\left|b\right|}\sqrt{b^{2} + x^{2}}$ approaches $\pm \frac{ax}{\left|b\right|}$ asymptotically. We can use the asymptote equations to calculate the values of $a$ and $b.$ Additionally, we can take the center of the center of the hyperbola as the point where its asymptotes intersect.
• To find the center of the hyperbola, we find the point at which its asymptotes intersect. Setting $2x + 1 = -2x + 3$ we find $x = \frac{1}{2}$ and $y = 2x + 1 = 2.$ Thus, the standard form for our hyperbola becomes $$\frac{(y - 2)^{2}}{a^{2}} - \frac{(x - \frac{1}{2})^{2}}{b^{2}} = 1$$where $b^{2} = a^{2}(e^{2} - 1).$ Since we know the hyperbola passes through the origin, we can substitute $x = 0$ and $y = 0$ giving us $$\frac{4}{a^{2}} - \frac{1}{4b^{2}} = 1$$Given the value $y = \pm \frac{a}{\left|b\right|} \sqrt{b^{2} + x^{2}}$ approaches $\pm \frac{ax}{\left|b\right|}$ asymptotically, we see that $\frac{a}{b}$ approaches $\pm 2,$ thus $b^{2}$ approaches $\frac{a^{2}}{4},$ turning the above relation into $$\frac{3}{a^{2}} = 1$$or in other words, $a^{2} = 3$ and $b^{2} = \frac{a^{2}}{4} = \frac{3}{4}.$ Plugging this back into our standard equation, we get $$(y - 2)^{2} - 4(x - \frac{1}{2})^{2} = 3$$Expanding and rearranging terms, we get $$y^{2} - 4x^{2} - 4y + 4x = 0 \quad \blacksquare$$