[b]Classify the conic sections described by the following equations.[br]Then, compute their foci, directrix/directrices, and eccentricity:[br][br]1. x^2+9y^2=4.[br][/b]Ellipse of the form: [math]\frac{x^2}{9}+y^2=\frac{4}{9}[/math] simplified further into [math]\frac{x^2}{4}+\frac{y^2}{\frac{4}{9}}=1[/math] or [math]\frac{x^2}{2^2}+\frac{y^2}{\left(\frac{2}{3}\right)^2}=1[/math][b] .[/b][br]Center: (0,0)[br]Foci: [math]c^2=2^2-\left(\frac{2}{3}\right)^2[/math] so [math]c=\frac{4\sqrt{2}}{3}[/math] and the foci are [math]\left(-\frac{4\sqrt{2}}{3},0\right)and\left(\frac{4\sqrt{2}}{3},0\right)[/math][br]Eccentricity = c/a. So e=[math]\frac{4\sqrt{2}}{3}\div2[/math] or [math]\frac{2\sqrt{2}}{3}[/math][br][br][br][b]2. 4x^2−y^2=9.[/b][br]Hyperbola of the form: [math]\frac{x^2}{\left(\frac{3}{2}\right)^2}-\frac{y^2}{3^2}=1[/math][br]Center: (0,0)[br]Foci: [math]\left(-\frac{3\sqrt{5}}{2},0\right)and\left(\frac{3\sqrt{5}}{2},0\right)[/math] and latus rectum at x = [math]-\frac{3\sqrt{5}}{2}[/math] and x = [math]\frac{3\sqrt{5}}{2}[/math][br]Eccentricity: e=[math]\frac{3\sqrt{5}}{2}\div\frac{3}{2}[/math] which is [math]\sqrt{5}[/math][br][br][br][b]3. x^2+4xy+4y^2+6x+5y−1=0.[/b][br][br]Slanted parabola of the form: