Q.Identify the type of conic and find centre, foci, vertices, and directrices of each of the following:

Part (i). Ellipse, $289>225$ (major axis vertical). $a^2=289,b^2=225\Rightarrow a=17,b=15$. $c^2=289-225=64\Rightarrow c=8$. Centre $(3,4)$; foci $(3,4\pm8)=(3,12),(3,-4)$; vertices $(3,4\pm17)=(3,21),(3,-13)$; directrices $y=4\pm\frac{289}8$.

Part (ii). Ellipse, $100>64$ (major axis horizontal). $a^2=100,b^2=64\Rightarrow a=10,b=8$. $c^2=100-64=36\Rightarrow c=6$. Centre $(-1,2)$; foci $(-1\pm6,2)=(5,2),(-7,2)$; vertices $(-1\pm10,2)=(9,2),(-11,2)$; directrices $x=-1\pm\frac{100}6$.

Part (iii). Hyperbola, positive $x$-term (transverse axis horizontal). $a^2=225,b^2=64\Rightarrow a=15,b=8$. $c^2=225+64=289\Rightarrow c=17$. Centre $(-3,4)$; foci $(-3\pm17,4)=(14,4),(-20,4)$; vertices $(-3\pm15,4)=(12,4),(-18,4)$; directrices $x=-3\pm\frac{225}{17}$.

Part (iv). Hyperbola, positive $y$-term (transverse axis vertical). $a^2=25,b^2=16\Rightarrow a=5,b=4$. $c^2=25+16=41\Rightarrow c=\sqrt{41}$. Centre $(-1,2)$; foci $(-1,2\pm\sqrt{41})$; vertices $(-1,2\pm5)=(-1,7),(-1,-3)$; directrices $y=2\pm\frac{25}{\sqrt{41}}$.

Part (v). $18x^2+12y^2-144x+48y+120=0$. Divide by $6$: $3x^2+2y^2-24x+8y+20=0$. Complete the square: $3(x^2-8x)+2(y^2+4y)+20=0 \Rightarrow 3(x-4)^2-48+2(y+2)^2-8+20=0 \Rightarrow 3(x-4)^2+2(y+2)^2=36$. Divide by $36$: $\frac{(x-4)^2}{12}+\frac{(y+2)^2}{18}=1$ — ellipse, $18>12$ (major axis vertical). $a^2=18,b^2=12$. $c^2=18-12=6\Rightarrow c=\sqrt6$. Centre $(4,-2)$; foci $(4,-2\pm\sqrt6)$; vertices $(4,-2\pm3\sqrt2)$ (since $a=\sqrt{18}=3\sqrt2$); directrices $y=-2\pm\frac{18}{\sqrt6}=-2\pm3\sqrt6$. …

Read a shifted standard form directly, or complete the square (both variables) …

• Dividing by the wrong (or forgetting to divide by a) leading coefficient in (v)/(vi) before completing the square …