Conic Sections — Free Pre-Calculus Review Games.

\(x^2 + y^2 = r^2\) is a circle with radius 5.

Standard form (x-h)^2+(y-k)^2=r^2 has center (h,k) = (2,-3).

A parabola has exactly one focus point.

\(r^2 = 49\), \(r = 7\).

An ellipse has exactly two foci.

Different denominators under \(x^2\) and \(y^2\) with + sign: ellipse.

Minus sign between squared terms: hyperbola.

Vertex form y = a(x-h)^2+k, vertex is (h,k) = (1,3).

\(a^2 = 25\), \(a = 5\). Major axis length = \(2a = 10\).

Vertical parabola: \((x-h)^2 = 4p(y-k)\) where \(p > 0\) opens up.

\(c^2 = a^2-b^2 = 25-9 = 16\), \(c = 4\). Major axis horizontal: foci at \((\pm4, 0)\).

Asymptotes: \(y = \pm(b/a)x = \pm(3/4)x\).

Standard form with h=3, k=-1, r=6: (x-3)^2+(y+1)^2=36.

A circle has eccentricity 0 (both foci are at the center).

A hyperbola always has eccentricity greater than 1.

For a horizontal major axis, the larger denominator \(a^2\) must be under \(x^2\), since a>b makes the ellipse wider than tall. The choice "\(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) where \(b>a>0\)" is wrong because if b>a the ellipse would be taller than wide, giving a vertical major axis instead. Always identify which denominator is larger to determine whether an ellipse's major axis runs horizontally or vertically.

A parabola is defined as the set of points equidistant from a fixed point called the focus and a fixed line called the directrix, which generates its curved shape. The choice "the two foci" is wrong because a parabola has only one focus, not two; having two foci is instead the defining property of ellipses and hyperbolas. Remembering this focus-directrix definition helps distinguish parabolas from the other conic sections, which are defined using two fixed points.

When the transverse axis is vertical, the positive squared term must be \(y^2\), so the hyperbola opens upward and downward along the y-axis. The equation "\(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\)" is wrong because it places the positive term under \(x^2\), producing a hyperbola that opens left and right instead. The sign and position of the positive term tell you both the type of conic and the direction it opens.

An ellipse has exactly two vertices, located at the endpoints of the major axis, while the endpoints of the minor axis are called co-vertices rather than vertices. The choice "4" confuses vertices with the total number of axis endpoints, since ellipses have four axis endpoints but only two are formally named vertices. Keep the vocabulary distinct: vertices sit on the major axis, and co-vertices sit on the minor axis.

Since \(y^2=4px\) with \(4p=8\) gives \(p=2>0\), and \(y^2\) equations open horizontally, a positive p means the parabola opens to the right. The choice "upward" is wrong because equations of the form \(y^2=4px\) always open left or right, while \(x^2=4py\) forms open up or down. Match the squared variable to the axis of opening: \(y^2\) terms open horizontally and \(x^2\) terms open vertically.

The defining property of an ellipse is that for any point on the curve, the sum of its distances to the two foci remains the same constant value, equal to the length of the major axis. The option "zero" is wrong because distances are always non-negative and cannot sum to zero unless the point coincides with both foci, which is impossible for distinct foci. This constant-sum property is what allows the "string and two tacks" method to draw an ellipse.

A hyperbola consists of points where the absolute value of the difference between distances to the two foci is always the same constant, unlike an ellipse where the sum is constant. The choice "the sum of distances to the foci" is wrong because that describes an ellipse, not a hyperbola. Contrasting "sum" for ellipses with "difference" for hyperbolas is a quick way to distinguish the two conics.

The general second-degree equation $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$ can represent a circle, ellipse, parabola, or hyperbola depending on the values of A, B, and C. The option "\(Ax+By+C=0\)" is wrong because it is a first-degree equation that only describes a straight line, not a curved conic section. Recognizing this general quadratic form is essential for identifying which conic an equation represents before converting it to standard form.

An ellipse always has an eccentricity strictly between 0 and 1, since \(e=c/a\) with \(c<a\) for a bounded, oval-shaped curve. The choice "\(e=0\)" is wrong because an eccentricity of exactly zero describes a circle, a special case where the foci merge into a single center point. Eccentricity closer to 0 means a more circular ellipse, while values closer to 1 mean a more elongated one.

Every parabola has an eccentricity of exactly 1, the boundary value separating ellipses (eccentricity less than 1) from hyperbolas (eccentricity greater than 1). The option "between 0 and 1" is wrong because that range describes ellipses, not parabolas, which sit exactly at the value 1. Memorizing e=1 for parabolas gives you a fast way to classify a conic once its eccentricity is known.

A hyperbola always has eccentricity greater than 1 because \(c>a\), meaning the foci lie farther from the center than the vertices do. The choice "it always equals 1" is wrong since an eccentricity of exactly 1 belongs to parabolas, not hyperbolas. Use e>1 as a quick check that a given equation truly describes a hyperbola rather than another conic.

A parabola opening to the right has its axis of symmetry along the x-axis and is written as \(y^2=4px\) with a positive p value, since positive p shifts the focus to the right of the vertex. The form "\(x^2=4py\) with \(p>0\)" is wrong because that equation describes a parabola opening upward along a vertical axis, not to the right. Always check both the squared variable and the sign of p to determine a parabola's orientation.

The minor axis of an ellipse runs along the shorter semi-axis, so its total length is twice the smaller value, \(2b\), since it stretches from \(-b\) to \(b\). The choice "\(2a\)" is wrong because that quantity gives the length of the major axis, which runs along the larger semi-axis \(a\). Always double the semi-axis length, whether major or minor, to get the full axis length.

Comparing \(x^2=8y\) to \(x^2=4py\) gives \(4p=8\), so \(p=2\), and since the parabola opens upward the focus lies at \((0,p)=(0,2)\). The point "\((2,0)\)" is wrong because it swaps the coordinates, which would only apply to a horizontally opening parabola of the form \(y^2=4px\). Always match the form of the equation to the correct axis before plugging in the value of p.

Matching \(x^2=4py\) gives \(4p=-12\) so \(p=-3\), and the directrix of a vertical parabola is the horizontal line \(y=-p\), which here is \(y=3\). The choice "\(y=-3\)" is wrong because it places the directrix on the same side as the focus rather than on the opposite side of the vertex. Remember that the directrix is always on the opposite side of the vertex from the focus and opening direction.

Completing the square on \(x=2y^2-4y+3\) gives \(x=2(y-1)^2+1\), so the vertex, where the parabola turns, is at \((1,1)\). The point "\((1,-1)\)" is wrong because it uses the incorrect sign for the y-coordinate obtained from completing the square inside the parentheses. Completing the square converts a general parabola equation into vertex form, directly revealing the vertex coordinates.

Since \(a^2=49\) gives \(a=7\) as the semi-major axis, the full major axis length is \(2a=14\). The choice "7" is wrong because it only represents the semi-major axis, not the full length of the axis from one end to the other. Always double the semi-axis value to report the full axis length rather than the radius-like semi-axis.

Because 49 is the larger denominator and sits under \(y^2\), the major axis is vertical with \(a=7\), while \(b^2=25\) gives \(b=5\) as the semi-minor axis, making the full minor axis length \(2b=10\). The choice "14" is wrong because that is the length of the major axis, not the minor axis. Always identify which denominator is larger first, since that determines which axis is major versus minor.

With \(a^2=16\) and \(b^2=9\), the focal distance is \(c=\sqrt{a^2-b^2}=\sqrt{7}\), so the eccentricity is \(e=c/a=\frac{\sqrt{7}}{4}\). The choice "\(\frac{3}{4}\)" is wrong because it uses \(b\) instead of \(c\) in the numerator, ignoring the correct formula relating a, b, and c. Always compute \(c=\sqrt{a^2-b^2}\) first for an ellipse before finding eccentricity as \(e=c/a\).

In standard form \(\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1\), the center is at \((h,k)\), so reading off \(h=1\) and \(k=-2\) gives the center \((1,-2)\). The choice "\((-1,2)\)" is wrong because it reverses the signs, which happens if you forget the equation subtracts h and k rather than adds them. Always solve \(x-h=0\) and \(y-k=0\) carefully to extract the true center coordinates from shifted conic equations.

Since the positive term is under \(y^2\) with \(a^2=4\), the transverse axis is vertical and the vertices lie at \((0,\pm a)=(0,\pm2)\). The choice "\((2,0)\) and \((-2,0)\)" is wrong because it places the vertices on the x-axis, which is only correct if the positive term were under \(x^2\). Always check which variable carries the positive term to determine whether the vertices lie on the x-axis or y-axis.

Completing the square on the y-terms gives \((y-2)^2=8x\), a horizontal parabola in vertex form \((y-k)^2=4p(x-h)\), revealing the vertex at \((h,k)=(0,2)\). The point "\((2,0)\)" is wrong because it swaps the roles of h and k, misreading which coordinate came from completing the square. When a parabola equation has a squared y-term, expect the vertex form \((y-k)^2=4p(x-h)\) with h and k read directly from inside the parentheses.

Completing the square on both variables reduces the equation to \(\frac{(x-2)^2}{9}+\frac{(y+1)^2}{4}=1\), the standard form of an ellipse since both squared terms are positive and added with different denominators. The choice "circle" is wrong because a circle requires equal denominators after normalizing, but here the denominators 9 and 4 differ. When both squared terms have the same sign but different coefficients, the conic is an ellipse rather than a circle.

Completing the square gives \((x-1)^2-(y-2)^2=1\), where the two squared terms have opposite signs, which is the defining characteristic of a hyperbola. The choice "ellipse" is wrong because ellipses require both squared terms to have the same sign, while here one term is subtracted from the other. A quick way to classify a conic is to check the signs of the squared x and y terms: same sign means ellipse or circle, opposite signs mean hyperbola.

With vertices at \((\pm5,0)\), \(a=5\), and foci at \((\pm3,0)\), \(c=3\), so \(b^2=a^2-c^2=25-9=16\), giving \(\frac{x^2}{25}+\frac{y^2}{16}=1\). The choice "\(\frac{x^2}{25}+\frac{y^2}{9}=1\)" is wrong because it mistakenly uses \(c^2=9\) in place of \(b^2\), confusing the focal distance with the semi-minor axis. Always apply \(b^2=a^2-c^2\) correctly to convert given vertex and focus information into the semi-minor axis value.

Since the vertex is at the origin and the focus is above it at \((0,3)\), the parabola opens upward with \(p=3\), giving \(x^2=4py=12y\). The choice "\(y^2=12x\)" is wrong because that form describes a parabola opening horizontally, placing the focus on the x-axis instead of the y-axis. Match the location of the focus relative to the vertex to choose the correct squared variable and the correct value of p.

The length of the latus rectum, the chord through the focus perpendicular to the axis, equals \(|4p|\), and since \(4p=12\) directly from the equation, the latus rectum length is 12. The choice "3" is wrong because that is only the value of p, the distance from vertex to focus, not the full chord length through the focus. Recall that the latus rectum length equals \(4p\) for any parabola in standard form, giving a quick way to describe the curve's width at the focus.

The transverse axis length of a hyperbola equals \(2a\), and since \(a^2=36\) gives \(a=6\), the transverse axis length is \(2(6)=12\). The choice "16" is wrong because that value relates to \(2b=2\sqrt{64}=16\), the conjugate axis length, not the transverse axis. Always use the denominator under the positive squared term to find \(a\) and double it for the transverse axis length.

The conjugate axis of a hyperbola has length \(2b\), and here \(b^2=16\) gives \(b=4\), so the conjugate axis length is \(2(4)=8\). The choice "10" is wrong because that value comes from doubling \(a=5\) from \(a^2=25\), which gives the transverse axis length instead. Remember that the denominator under the negative term corresponds to \(b\) and determines the conjugate axis, perpendicular to the transverse axis.

The focal distance is \(c=\sqrt{a^2-b^2}=\sqrt{64-36}=\sqrt{28}=2\sqrt{7}\), and since the two foci are symmetric about the center, the total distance between them is \(2c=4\sqrt{7}\). The choice "\(2\sqrt{7}\)" is wrong because it represents only the distance from the center to one focus, not the full distance between both foci. Always double the value of c to report the distance between two foci rather than leaving it as a single focal radius.

For a hyperbola, \(c^2=a^2+b^2=9+16=25\), so \(c=5\), and since the transverse axis is horizontal, the foci lie at \((\pm5,0)\). The choice "\((0,\pm5)\)" is wrong because it places the foci on the y-axis, which would only be correct for a hyperbola with a vertical transverse axis. Remember that hyperbolas use \(c^2=a^2+b^2\), unlike ellipses which use \(c^2=a^2-b^2\), so never mix up the two formulas.

When \(B^2-4AC\) equals exactly zero, the general second-degree equation represents a parabola, the boundary between the ellipse case (negative discriminant) and the hyperbola case (positive discriminant). The choice "the conic is a hyperbola" is wrong because a hyperbola corresponds to a positive discriminant, \(B^2-4AC>0\), not zero. Memorize the discriminant test — negative gives ellipse or circle, zero gives parabola, positive gives hyperbola — as a fast classification tool for any general conic equation.

Since \(a=4\) from the vertices and \(c=6\) from the foci, the relation \(b^2=c^2-a^2=36-16=20\) gives the equation \(\frac{x^2}{16}-\frac{y^2}{20}=1\). The choice "\(\frac{x^2}{20}-\frac{y^2}{16}=1\)" is wrong because it swaps the values of \(a^2\) and \(b^2\), incorrectly changing the vertex locations to \((\pm\sqrt{20},0)\). Always apply \(b^2=c^2-a^2\) for hyperbolas, the opposite relationship from ellipses, and keep \(a^2\) under the term matching the transverse axis.

Since eccentricity is defined as \(e=c/a\), rearranging gives \(a=c/e=4\div\frac{2}{3}=6\), the semi-major axis length. The choice "\(8\)" is wrong because it results from multiplying c by e instead of dividing, reversing the correct algebraic operation. When given eccentricity and c, always solve for a using \(a=c/e\) rather than multiplying the two values together.

Since the directrix \(x=-4\) is vertical, the parabola's axis is horizontal, so the vertex lies midway between the focus and directrix, giving x-coordinate \(\frac{2+(-4)}{2}=-1\) and the same y-coordinate as the focus, 3, so the vertex is \((-1,3)\). The choice "\((-1,0)\)" is wrong because it drops the y-coordinate to zero instead of matching the focus's y-value, which must stay the same along a horizontal axis of symmetry. The vertex always lies exactly halfway between the focus and directrix, measured along the parabola's axis of symmetry.

For a hyperbola, \(c=\sqrt{a^2+b^2}=\sqrt{9+16}=5\), so the eccentricity is \(e=c/a=\frac{5}{3}\), a value greater than 1 as expected. The choice "\(\frac{3}{5}\)" is wrong because it inverts the ratio, computing \(a/c\) instead of \(c/a\), which would give an eccentricity less than 1, impossible for a hyperbola. Always compute c first using \(c^2=a^2+b^2\) before forming the eccentricity ratio \(e=c/a\).

Here \(A=4\), \(B=0\), and \(C=-1\), so \(B^2-4AC=0-4(4)(-1)=16\), a positive value confirming the equation is a hyperbola. The choice "a circle, since \(A=C\)" is wrong because \(A=4\) and \(C=-1\) are not equal, and even if they were, opposite signs on the squared terms would still rule out a circle. Always compute the discriminant from the general form's coefficients A, B, and C rather than relying on appearance alone to classify a conic.

By the defining property of an ellipse, the sum of distances from any point on the curve to the two foci equals \(2a=2(5)=10\), so if one distance is 6, the other must be \(10-6=4\). The choice "\(10\)" is wrong because that is the total constant sum, not the remaining individual distance after subtracting the known 6. Always use \(2a\) as the fixed total sum, then subtract the known distance to isolate the unknown one.

The semi-major axis is \(a=\frac{2+8}{2}=5\) AU, and since perihelion equals \(a-c\), solving \(2=5-c\) gives \(c=3\), so the eccentricity is \(e=c/a=3/5=0.6\). The choice "\(0.5\)" is wrong because it does not correctly account for the asymmetry between perihelion and aphelion distances relative to the center. For orbital problems, remember that perihelion equals \(a-c\) and aphelion equals \(a+c\), letting you solve for both a and c from the two extreme distances.

Regardless of which point on the ellipse the sound reflects off, the total path length from one focus to the wall and back to the other focus always equals the constant sum of distances, \(2a\), which equals the major axis length of 40 feet. The choice "\(24\) feet" is wrong because that value is the distance between the two foci, \(2c\), not the constant sum of distances that defines the ellipse. This reflective property, where any path between foci has the same total length, is exactly why whispering galleries work acoustically at any point along the curved wall.

The constant difference in distances to the two foci equals \(2a\) by the defining property of a hyperbola, so setting \(2a=40\) gives \(a=20\). The choice "\(50\)" is wrong because that is half the distance between the towers, \(c\), which relates to the foci separation rather than the constant distance difference. In navigation applications like this, always equate the given constant distance difference to \(2a\), not to the distance between the foci, which equals \(2c\).

Using \(e^2=1+\left(\frac{b}{a}\right)^2\), substituting \(e=\frac{5}{4}\) gives \(\frac{25}{16}=1+\left(\frac{b}{a}\right)^2\), so \(\left(\frac{b}{a}\right)^2=\frac{9}{16}\) and \(\frac{b}{a}=\frac{3}{4}\), the slope of the asymptotes \(y=\pm\frac{b}{a}x\). The choice "\(\pm\frac{5}{4}\)" is wrong because that value is the eccentricity itself, not the ratio \(b/a\) obtained after solving the eccentricity relation. Remember the identity \(e^2=1+(b/a)^2\) for hyperbolas connects eccentricity directly to the asymptote slopes, a useful shortcut on multi-step problems.

Whenever a general conic equation contains a nonzero \(B\) coefficient on the \(xy\) term, the axes must be rotated by an appropriate angle to eliminate that cross term before the equation can be written in a recognizable standard form. The choice "complete the square only, since no rotation is needed" is wrong because completing the square cannot remove a cross term like \(xy\); only a rotation transformation accomplishes that. Whenever B is nonzero in the general conic equation, expect the curve to be tilted relative to the coordinate axes, requiring rotation before further classification.

For a downward-opening parabola with vertex \((3,-2)\) and \(|p|=1\), the focus lies below the vertex at \((3,-3)\), and the directrix lies the same distance above the vertex on the opposite side, giving the horizontal line \(y=-1\). The choice "\(y=-3\)" is wrong because that is actually the location of the focus, not the directrix, which must be the same distance from the vertex but in the opposite direction. Always place the focus and directrix on opposite sides of the vertex, each at distance \(|p|\), to avoid mixing up their locations.

Substituting a line's equation into the ellipse's equation produces a quadratic equation in one variable, and a quadratic can have at most two real roots, meaning a line can intersect an ellipse in at most two points. The choice "4, because an ellipse has two axes each line can cross" is wrong because intersection count depends on the degree of the resulting equation, not on the number of axes the ellipse happens to have. This algebraic reasoning, substitute and count the degree of the resulting polynomial, applies generally to finding intersections between lines and any conic section.