Solution.

(a)

Write it in standard form: \(x^2 = -\dfrac {7}{5}y\). Comparing with \(x^2 = -4py\) gives \(4p = \dfrac {7}{5}\), so \(p = \dfrac {7}{20}\). The parabola opens downward with vertex at the origin, focus \(\Big (0, -\dfrac {7}{20}\Big )\) and directrix \(y = \dfrac {7}{20}\).

(b)

Here \(y^2 = -8x\). Comparing with \(y^2 = -4px\) gives \(4p = 8\), so \(p = 2\). The parabola opens to the left with vertex at the origin, focus \((-2, 0)\) and directrix \(x = 2\).

Solution. A point \(P(x,y)\) lies on the parabola when its distance to the focus equals its distance to the directrix: \[\sqrt {(x-4)^2 + (y-5)^2} = \left |y + 1\right | .\] Squaring, \[x^2 - 8x + 16 + y^2 - 10y + 25 = y^2 + 2y + 1 \hspace {0.4cm}\implies \hspace {0.4cm} x^2 - 8x + 40 = 12y .\] Completing the square, \((x-4)^2 = 12(y-2)\). The vertex is \((4,2)\), midway between the focus and the directrix as it must be, and \(4p = 12\) so \(p = 3\).

Solution. The distance sum is \(2a\) and the foci are at distance \(c\) from the centre, with \(b^2 = a^2 - c^2\).

(a)

\(2a = 10\) so \(a = 5\), and \(c = 3\), giving \(b^2 = 25 - 9 = 16\). The foci are on the \(x-\)axis, so \[\frac {x^2}{25} + \frac {y^2}{16} = 1 .\]

(b)

\(2a = 20\) so \(a = 10\), and \(c = 6\), giving \(b^2 = 100 - 36 = 64\). The foci are on the \(y-\)axis, so the larger denominator belongs to \(y\): \[\frac {x^2}{64} + \frac {y^2}{100} = 1 .\]

Solution.

(a)

The larger denominator is under \(x\), so \(a^2 = 25\), \(b^2 = 9\) and the major axis is horizontal. Then \(c^2 = 25 - 9 = 16\), so \(c = 4\): foci \((\pm 4, 0)\) and distance sum \(2a = 10\).

(b)

The same numbers with the roles of \(x\) and \(y\) exchanged: \(a^2 = 25\) under \(y\), so the major axis is vertical, \(c = 4\), foci \((0, \pm 4)\) and distance sum \(10\).

Solution. Take the ellipse centred at the origin with axes along the coordinate axes, \[\frac {x^2}{A} + \frac {y^2}{B} = 1 .\] Substituting the two points gives \[\frac {16}{A} + \frac {1}{B} = 1, \qquad \frac {4}{A} + \frac {4}{B} = 1 .\] Multiply the first by \(4\) and subtract the second: \[\frac {64}{A} - \frac {4}{A} = 3 \implies \frac {60}{A} = 3 \implies A = 20 .\] Then \(\dfrac {16}{20} + \dfrac {1}{B} = 1\) gives \(\dfrac {1}{B} = \dfrac {1}{5}\), so \(B = 5\). Hence \[\frac {x^2}{20} + \frac {y^2}{5} = 1 .\]

Solution. The distance difference is \(2a\) and \(b^2 = c^2 - a^2\).

(a)

\(2a = 8\) so \(a = 4\), and \(c = 5\), giving \(b^2 = 25 - 16 = 9\). The foci are on the \(x-\)axis, so \[\frac {x^2}{16} - \frac {y^2}{9} = 1 .\]

(b)

The same values with the foci on the \(y-\)axis, so the positive term is in \(y\): \[\frac {y^2}{16} - \frac {x^2}{9} = 1 .\]

Solution. For a hyperbola it is the sign, not the size of the denominators, that fixes the axis, and \(c^2 = a^2 + b^2\) in both cases with \(a^2\) under the positive term.

(a)

\(a^2 = 25\), \(b^2 = 144\), so \(c^2 = 169\) and \(c = 13\): foci \((\pm 13, 0)\) and distance difference \(2a = 10\).

(b)

The positive term is in \(y\), so the curve opens vertically: foci \((0, \pm 13)\) and distance difference \(10\).

Solution. The foci are on the \(x-\)axis with \(c = 2\), so \(a^2 + b^2 = 4\) and \(b^2 = 4 - a^2\). Substituting the point \((2,3)\) into \(\dfrac {x^2}{a^2} - \dfrac {y^2}{b^2} = 1\), \[\frac {4}{a^2} - \frac {9}{4-a^2} = 1 .\] Multiplying by \(a^2(4-a^2)\), \[16 - 13a^2 = 4a^2 - a^4 \implies a^4 - 17a^2 + 16 = 0 \implies \big (a^2-1\big )\big (a^2-16\big ) = 0 .\] Since \(a^2 < c^2 = 4\), we take \(a^2 = 1\) and \(b^2 = 3\): \[x^2 - \frac {y^2}{3} = 1 ,\] which checks: \(4 - \dfrac {9}{3} = 1\).

Solution. \[9\big (x^2-10x\big ) + 25\big (y^2-6y\big ) = -225\] \[9(x-5)^2 - 225 + 25(y-3)^2 - 225 = -225 \implies 9(x-5)^2 + 25(y-3)^2 = 225 .\] Dividing by \(225\), \[\frac {(x-5)^2}{25} + \frac {(y-3)^2}{9} = 1 .\] An ellipse, centre \((5,3)\), \(a = 5\), \(b = 3\), \(c = 4\). Major axis horizontal, vertices \((0,3)\) and \((10,3)\), foci \((1,3)\) and \((9,3)\), \(e = \dfrac {4}{5}\).

Solution. \[16(y-2)^2 - 64 - 9(x+3)^2 + 81 = 161 \implies 16(y-2)^2 - 9(x+3)^2 = 144 ,\] \[\frac {(y-2)^2}{9} - \frac {(x+3)^2}{16} = 1 .\] A hyperbola, centre \((-3,2)\), opening vertically since the positive term is in \(y\). Here \(a = 3\), \(b = 4\), \(c = 5\): vertices \((-3, 2\pm 3)\), foci \((-3, 2\pm 5)\), asymptotes \(y - 2 = \pm \dfrac {3}{4}(x+3)\), \(e = \dfrac {5}{3}\).

Solution. Only \(x\) is squared, so this is a parabola. \[(x+1)^2 - 1 + 16y + 33 = 0 \implies (x+1)^2 = -16(y+2) .\] Vertex \((-1,-2)\), opening downward, \(4p = 16\) so \(p = 4\): focus \((-1,-6)\) and directrix \(y = 2\).

Solution. \[16(x-1)^2 - 16 + (y+2)^2 - 4 + 16 = 0 \implies 16(x-1)^2 + (y+2)^2 = 4 ,\] \[\frac {(x-1)^2}{\frac {1}{4}} + \frac {(y+2)^2}{4} = 1 .\] An ellipse, centre \((1,-2)\), with \(a^2 = 4\) under \(y\) so the major axis is vertical: \(a = 2\), \(b = \dfrac {1}{2}\), and \(c = \sqrt {4 - \frac 14} = \dfrac {\sqrt {15}}{2}\). Foci \(\Big (1, -2 \pm \dfrac {\sqrt {15}}{2}\Big )\) and \(e = \dfrac {\sqrt {15}}{4}\).

Solution. Only \(y\) is squared, so this is a parabola. \[20(y+1)^2 - 20 + x + 27 = 0 \implies 20(y+1)^2 = -(x+7) \implies (y+1)^2 = -\frac {1}{20}(x+7).\] Vertex \((-7,-1)\), opening to the left, with \(4p = \dfrac {1}{20}\) so \(p = \dfrac {1}{80}\): focus \(\Big (-7 - \dfrac {1}{80},\, -1\Big )\) and directrix \(x = -7 + \dfrac {1}{80}\).

Solution. Bring everything to one side: \(y^2 - 10y - x^2 - 4x + 20 = 0\), so \[(y-5)^2 - 25 - \big [(x+2)^2 - 4\big ] + 20 = 0 \implies (y-5)^2 - (x+2)^2 = 1 .\] A hyperbola, centre \((-2,5)\), opening vertically, with \(a = b = 1\) and \(c = \sqrt 2\). Vertices \((-2, 5\pm 1)\), foci \((-2, 5\pm \sqrt 2)\), asymptotes \(y - 5 = \pm (x+2)\), and \(e = \sqrt 2\). Since \(a = b\) this is a rectangular hyperbola, and its asymptotes are perpendicular.

Solution. \[25(x+2)^2 - 100 + 16(y-6)^2 - 576 + 276 = 0 \implies 25(x+2)^2 + 16(y-6)^2 = 400 ,\] \[\frac {(x+2)^2}{16} + \frac {(y-6)^2}{25} = 1 .\] An ellipse, centre \((-2,6)\), major axis vertical with \(a = 5\), \(b = 4\) and \(c = 3\): vertices \((-2, 6\pm 5)\), foci \((-2, 6\pm 3)\), \(e = \dfrac {3}{5}\).

Solution. \[6\big (x^2+14x\big ) - 15\big (y^2+6y\big ) + 69 = 0\] \[6(x+7)^2 - 294 - 15(y+3)^2 + 135 + 69 = 0 \implies 6(x+7)^2 - 15(y+3)^2 = 90 ,\] \[\frac {(x+7)^2}{15} - \frac {(y+3)^2}{6} = 1 .\] A hyperbola, centre \((-7,-3)\), opening horizontally, \(a^2 = 15\), \(b^2 = 6\), \(c = \sqrt {21}\). Asymptotes \(y + 3 = \pm \sqrt {\dfrac {6}{15}}(x+7)\) and \(e = \sqrt {\dfrac {21}{15}} = \dfrac {\sqrt {35}}{5}\).

Solution. Only \(x\) is squared, so this is a parabola. Rearranging, \[x^2 - 18x + 7950 - 2y = 0 \implies (x-9)^2 - 81 + 7950 = 2y \implies (x-9)^2 = 2\Big (y - \frac {7869}{2}\Big ).\] Vertex \(\Big (9, \dfrac {7869}{2}\Big )\), opening upward, with \(4p = 2\) so \(p = \dfrac {1}{2}\): focus \(\Big (9, \dfrac {7870}{2}\Big ) = (9, 3935)\) and directrix \(y = \dfrac {7868}{2} = 3934\).

Solution. \[3x^2 + 4(y-2)^2 - 16 = 92 \implies 3x^2 + 4(y-2)^2 = 108 ,\] \[\frac {x^2}{36} + \frac {(y-2)^2}{27} = 1 .\] An ellipse with centre \((0,2)\), \(a^2 = 36\) under \(x\) so the major axis is horizontal, \(a = 6\), \(b^2 = 27\), and \[c = \sqrt {36-27} = 3, \qquad e = \frac {c}{a} = \frac {1}{2}.\] Foci \((\pm 3, 2)\); directrices \(x = \pm \dfrac {a}{e} = \pm 12\).

Solution. \[25(x+4)^2 - 400 + 16y^2 + 400 = 160 \implies 25(x+4)^2 + 16y^2 = 160 ,\] \[\frac {(x+4)^2}{\frac {32}{5}} + \frac {y^2}{10} = 1 .\] An ellipse with centre \((-4, 0)\). The larger denominator is under \(y\), so the major axis is vertical with \(a^2 = 10\) and \(b^2 = \dfrac {32}{5}\). Then \[c^2 = 10 - \frac {32}{5} = \frac {18}{5} \implies c = \frac {3\sqrt {10}}{5}, \qquad e = \frac {c}{a} = \frac {3\sqrt {10}}{5\sqrt {10}} = \frac {3}{5}.\] Foci \(\Big (-4, \pm \dfrac {3\sqrt {10}}{5}\Big )\); directrices \(y = \pm \dfrac {a}{e} = \pm \dfrac {5\sqrt {10}}{3}\).

Solution. Expanding the right side and rearranging, \[8x^2 + 32x + 23 - y^2 - 2y = 0\] \[8(x+2)^2 - 32 - (y+1)^2 + 1 + 23 = 0 \implies 8(x+2)^2 - (y+1)^2 = 8 ,\] \[\frac {(x+2)^2}{1} - \frac {(y+1)^2}{8} = 1 .\] A hyperbola with centre \((-2,-1)\), opening horizontally, \(a = 1\), \(b^2 = 8\), \[c = \sqrt {1+8} = 3, \qquad e = \frac {c}{a} = 3 .\] Foci \((-2\pm 3, -1)\), that is \((1,-1)\) and \((-5,-1)\); directrices \(x = -2 \pm \dfrac {1}{3}\); asymptotes \(y + 1 = \pm 2\sqrt 2\,(x+2)\).

Solution. \[9y^2 - 72y - 16x^2 + 96x - 144 = 0\] \[9(y-4)^2 - 144 - 16(x-3)^2 + 144 - 144 = 0 \implies 9(y-4)^2 - 16(x-3)^2 = 144 ,\] \[\frac {(y-4)^2}{16} - \frac {(x-3)^2}{9} = 1 .\] A hyperbola with centre \((3,4)\), opening vertically, \(a = 4\), \(b = 3\), \[c = \sqrt {16+9} = 5, \qquad e = \frac {5}{4}.\] Foci \((3, 4\pm 5)\), that is \((3,9)\) and \((3,-1)\); directrices \(y = 4 \pm \dfrac {16}{5}\); asymptotes \(y - 4 = \pm \dfrac {4}{3}(x-3)\).

Solution. \(A = 1\), \(B = 2\), \(C = 1\), so \(B^2 - 4AC = 4 - 4 = 0\): a parabola. Since \(A = C\) the angle is \(\theta = \dfrac {\pi }{4}\), and with \(x = \dfrac {X-Y}{\sqrt 2}\), \(y = \dfrac {X+Y}{\sqrt 2}\) the equation becomes \[2X^2 - 8\sqrt 2\,Y = 0 \implies X^2 = 4\sqrt 2\,Y ,\] a parabola with vertex at the origin, axis along the \(Y-\)axis (the line \(y = x\)), and \(4p = 4\sqrt 2\) so \(p = \sqrt 2\).

Solution. \(B^2 - 4AC = 12 - 780 = -768 < 0\): an ellipse. Here \[\cot 2\theta = \frac {A-C}{B} = \frac {13-15}{-2\sqrt 3} = \frac {1}{\sqrt 3} \implies 2\theta = \frac {\pi }{3} \implies \theta = \frac {\pi }{6}.\] Substituting \(x = \dfrac {\sqrt 3 X - Y}{2}\), \(y = \dfrac {X + \sqrt 3 Y}{2}\), \[12X^2 + 16Y^2 = 192 \implies \frac {X^2}{16} + \frac {Y^2}{12} = 1 ,\] an ellipse with semi-axes \(4\) and \(2\sqrt 3\), the major axis along \(X\), at \(30^{\circ }\) to the \(x-\)axis.

Solution. \(B^2 - 4AC = 324 - 196 = 128 > 0\): a hyperbola. Since \(A = C\), \(\cot 2\theta = 0\) and \(\theta = \dfrac {\pi }{4}\). Substituting, \[-2X^2 + 16Y^2 = 24 \implies \frac {Y^2}{\frac {3}{2}} - \frac {X^2}{12} = 1 ,\] a hyperbola opening along the \(Y-\)axis, which is the line \(y = x\).

Solution. \(B^2 - 4AC = 36 + 60 = 96 > 0\): a hyperbola. Now \[\cot 2\theta = \frac {5-(-3)}{6} = \frac {4}{3} \implies \cos 2\theta = \frac {4}{5},\] so \(\cos \theta = \sqrt {\dfrac {1+\frac 45}{2}} = \dfrac {3}{\sqrt {10}}\) and \(\sin \theta = \sqrt {\dfrac {1-\frac 45}{2}} = \dfrac {1}{\sqrt {10}}\). Substituting, \[6X^2 - 4Y^2 = 24 \implies \frac {X^2}{4} - \frac {Y^2}{6} = 1 ,\] a hyperbola opening along the \(X-\)axis, at angle \(\theta = \tan ^{-1}\dfrac {1}{3} \approx 18.4^{\circ }\).

Solution. \(B^2 - 4AC = 108 - 364 = -256 < 0\): an ellipse. Here \[\cot 2\theta = \frac {7-13}{6\sqrt 3} = -\frac {1}{\sqrt 3} \implies 2\theta = \frac {2\pi }{3} \implies \theta = \frac {\pi }{3}.\] With \(\cos \theta = \dfrac 12\) and \(\sin \theta = \dfrac {\sqrt 3}{2}\), \[16X^2 + 4Y^2 = 64 \implies \frac {X^2}{4} + \frac {Y^2}{16} = 1 ,\] an ellipse with semi-axes \(2\) and \(4\), major axis along \(Y\), at \(60^{\circ }\) to the \(x-\)axis.

Solution. \(B^2 - 4AC = 576 - 576 = 0\): a parabola. Here \[\tan 2\theta = \frac {B}{A-C} = \frac {-24}{9-16} = \frac {24}{7} \implies \cos 2\theta = \frac {7}{25},\] so \(\cos \theta = \dfrac {4}{5}\) and \(\sin \theta = \dfrac {3}{5}\). Substituting, \[25Y^2 - 25X = 0 \implies Y^2 = X ,\] a parabola with vertex at the origin and axis along the \(X-\)axis, at \(\theta = \tan ^{-1}\dfrac 34 \approx 36.9^{\circ }\) to the \(x-\)axis, with \(4p = 1\) so \(p = \dfrac 14\).

Solution. \(B^2 - 4AC = 4 - 4 = 0\): a parabola. Since \(A = C\), \(\theta = \dfrac {\pi }{4}\), and substituting gives \[2Y^2 - 8\sqrt 2\,X + 6\sqrt 2\,Y + 33 = 0 .\] Completing the square in \(Y\), \[2\Big (Y + \frac {3\sqrt 2}{2}\Big )^2 = 8\sqrt 2\,X - 33 + 9 = 8\sqrt 2\Big (X - \frac {24}{8\sqrt 2}\Big ),\] a parabola with axis parallel to the \(X-\)axis. A translation to the new vertex puts it in the standard form \(Y'^2 = 4\sqrt 2\,X'\).

Solution. \(A = C = 0\) and \(B = 2\), so \(B^2 - 4AC = 4 > 0\): a hyperbola. Since \(A = C\) the angle is \(\theta = \dfrac {\pi }{4}\), and substituting gives \[X^2 - Y^2 - 5\sqrt 2\,X - \sqrt 2\,Y + 11 = 0 .\] Completing the square in each variable, \[\Big (X - \frac {5\sqrt 2}{2}\Big )^2 - \Big (Y + \frac {\sqrt 2}{2}\Big )^2 = -11 + \frac {25}{2} - \frac {1}{2} = 1 ,\] a rectangular hyperbola, centred at \(\Big (\dfrac {5\sqrt 2}{2}, -\dfrac {\sqrt 2}{2}\Big )\) in the rotated system, with \(a = b = 1\) and asymptotes at right angles.

Solution. The denominator already has constant term \(1\), so \(e = 1\) and \(ep = 7\), giving \(p = 7\). Since \(e = 1\) this is a parabola, with focus at the pole and directrix the horizontal line \(7\) units below it, that is \(y = -7\).

Solution. Here \(e = 3\) and \(ep = 5\), so \(p = \dfrac {5}{3}\). Since \(e > 1\) this is a hyperbola, with directrix the vertical line \(\dfrac {5}{3}\) units to the right of the pole.

Solution. Divide numerator and denominator by \(4\): \[r = \frac {\frac {10}{4}}{1 - \frac 34\cos \theta }.\] So \(e = \dfrac {3}{4}\) and \(ep = \dfrac {5}{2}\), giving \(p = \dfrac {5}{2}\cdot \dfrac {4}{3} = \dfrac {10}{3}\). Since \(e<1\) this is an ellipse, with directrix the vertical line \(\dfrac {10}{3}\) units to the left of the pole.

Solution. Dividing by \(3\), \[r = \frac {\frac {10}{3}}{1 + \frac 43\sin \theta },\] so \(e = \dfrac {4}{3}\) and \(ep = \dfrac {10}{3}\), giving \(p = \dfrac {10}{3}\cdot \dfrac {3}{4} = \dfrac {5}{2}\). Since \(e>1\) this is a hyperbola, with directrix the horizontal line \(\dfrac {5}{2}\) units above the pole.

Solution. Dividing by \(2\), \[r = \frac {6}{1 - \frac 12\sin \theta },\] so \(e = \dfrac {1}{2}\) and \(ep = 6\), giving \(p = 12\). Since \(e<1\) this is an ellipse, with directrix the horizontal line \(12\) units below the pole.