1. Problem 1: $\dfrac{x^2}{16} - \dfrac{y^2}{25} = 1$ - This is a hyperbola centered at the origin with transverse axis on the x-axis. - Here, $a^2 = 16$, so $a = 4$. - Also, $b^2 = 25$, so $b = 5$. - The foci satisfy $c^2 = a^2 + b^2 = 16 + 25 = 41$, so $c = \sqrt{41}$. Vertices (endpoints of transverse axis): $(\pm a, 0) = (\pm 4, 0)$. Foci: $(\pm c, 0) = (\pm \sqrt{41}, 0)$. Endpoints of conjugate axis: $(0, \pm b) = (0, \pm 5)$. Endpoints of latus rectum: For each focus $(\pm c, 0)$, the latus rectum endpoints are at $\left(\pm c, \pm \dfrac{b^2}{a}\right) = \left(\pm \sqrt{41}, \pm \dfrac{25}{4}\right)$. 2. Problem 2: $\dfrac{y^2}{9} - \dfrac{x^2}{7} = 1$ - Center at origin, transverse axis on the y-axis (since $y^2$ term is positive). - $a^2 = 9$, so $a=3$. - $b^2=7$, so $b=\sqrt{7}$. - $c^2 = a^2 + b^2 = 9 + 7=16$, so $c=4$. Vertices: $(0, \pm a) = (0, \pm 3)$. Foci: $(0, \pm c) = (0, \pm 4)$. Conjugate axis endpoints: $(\pm b, 0) = (\pm \sqrt{7}, 0)$. Latus rectum endpoints: For each focus $(0, \pm c)$, points are at $\left(\pm \dfrac{b^2}{a}, \pm c \right) = \left(\pm \dfrac{7}{3}, \pm 4 \right)$. 3. Problem 3: $y^2 - 25x^2 = 100$ - Rewrite in standard form: $\dfrac{y^2}{100} - \dfrac{x^2}{4} =1$. - $a^2=100$, so $a=10$. - $b^2=4$, so $b=2$. - $c^2 = a^2 + b^2 = 100 + 4 = 104$, so $c=2\sqrt{26}$. Vertices: $(0, \pm 10)$. Foci: $(0, \pm 2\sqrt{26})$. Conjugate axis endpoints: $(\pm 2, 0)$. Latus rectum endpoints: $\left(\pm \dfrac{b^2}{a}, \pm c \right) = \left(\pm \dfrac{4}{10}, \pm 2\sqrt{26}\right) = \left(\pm 0.4, \pm 2\sqrt{26}\right)$. 4. Problem 4: $9x^2 - 4y^2 - 36 =0$ - Rewrite: $9x^2 - 4y^2 = 36$ or $\dfrac{x^2}{4} - \dfrac{y^2}{9} =1$. - $a^2=4$, $a=2$. - $b^2=9$, $b=3$. - $c^2 = a^2 + b^2 = 4 + 9=13$, $c=\sqrt{13}$. Vertices: $(\pm 2, 0)$. Foci: $(\pm \sqrt{13}, 0)$. Conjugate axis endpoints: $(0, \pm 3)$. Latus rectum endpoints: $\left(\pm c, \pm \dfrac{b^2}{a} \right) = \left(\pm \sqrt{13}, \pm \dfrac{9}{2} \right)$.