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This deck focuses on Conic Sections, giving you a quick way to review the definitions, rules, and examples that matter most for AP Precalculus.
Study Conic Sections in AP Precalculus with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.
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Find the center of the circle (x−3)2+(y+5)2=36.
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(3, -5). The center is at (h,k) in the vertex form (x−h)2+(y−k)2=r2.
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This deck focuses on Conic Sections, giving you a quick way to review the definitions, rules, and examples that matter most for AP Precalculus.
Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.
Answer: (3, -5). The center is at (h,k) in the vertex form (x−h)2+(y−k)2=r2.
Answer: a in a2x2+b2y2=1. The larger denominator corresponds to the major axis length.
Answer: Hyperbola. Difference of squares with equal coefficients indicates a hyperbola.
Answer: Parabola. Square term on y with linear x indicates a horizontal parabola.
Answer: ∣4p∣. The chord through the focus perpendicular to the axis of symmetry.
Answer: A=πab. Where a and b are the lengths of the semi-major and semi-minor axes.
Answer: c=a2−b2. For an ellipse, c<a since the foci are inside the ellipse.
Answer: Circle. Complete the square to verify equal coefficients for x2 and y2.
Answer: Circle. Complete the square to verify it has equal coefficients for x2 and y2 terms.
Answer: Foci: (±5,0). From c2=a2+b2=9+16=25, so c=5.
Answer: Hyperbola. Divide by 36 to get standard form with a2=9 and b2=4.
Answer: A set of points equidistant from a fixed point (center). All points are the same distance from the center point.
Answer: Circle. Complete the square to verify it has equal coefficients for squared terms.
Answer: Ellipse. Divide by 144 to get standard form with positive coefficients.
Answer: c2=a2+b2. For hyperbolas, c>a since the foci are outside the vertices.
Answer: A set of points equidistant from a point (focus) and a line (directrix). This property defines the parabola's unique geometric shape.
Answer: (x2−x1)2+(y2−y1)2. Derived from the Pythagorean theorem in coordinate geometry.
Answer: (4, 0). From y2=16x, we get 4p=16, so p=4 and focus is at (p,0).
Answer: Circle. Complete the square to verify it forms a circle equation.
Answer: A set of points where the sum of distances to two foci is constant. The sum equals 2a, where a is the semi-major axis length.
Answer: A set of points where the sum of distances to two foci is constant. The sum equals 2a, where a is the semi-major axis length.
Answer: Ellipse. Divide by 144 to get standard form with positive coefficients.
Answer: ∣4p∣. The chord through the focus perpendicular to the axis of symmetry.
Answer: a2x2+b2y2=1. Where a and b are the semi-major and semi-minor axis lengths.
Answer: (2, 5). The vertex form shows the vertex at (h,k) where the parabola turns.
Answer: Hyperbola. Eccentricity greater than 1 distinguishes hyperbolas from other conics.
Answer: Circle. Equal coefficients (25) for x2 and y2 terms indicate a circle.
Answer: Hyperbola. Difference of squares form indicates a hyperbola with a2=b2=1.
Answer: (3, -5). The center is at (h,k) in the vertex form (x−h)2+(y−k)2=r2.
Answer: a2x2−b2y2=1. Subtraction between squared terms indicates a hyperbola opening horizontally.
Answer: Ax2+Bxy+Cy2+Dx+Ey+F=0. The discriminant B2−4AC determines the conic type.
Answer: A set of points where the difference of distances to two foci is constant. The absolute value of the difference equals 2a, where a is the semi-major axis.
Answer: Foci: (±5,0). From c2=a2+b2=9+16=25, so c=5.
Answer: A=πab. Where a and b are the lengths of the semi-major and semi-minor axes.
Answer: a2x2−b2y2=1. Subtraction between squared terms indicates a hyperbola opening horizontally.
Answer: Circle. Complete the square to verify it has equal coefficients for x2 and y2 terms.
Answer: (4, 0). From y2=16x, we get 4p=16, so p=4 and focus is at (p,0).
Answer: Circle. Complete the square to verify it forms a circle equation.
Answer: y=−2. From x2=8y, we have 4p=8, so p=2 and directrix is y=−p.
Answer: Parabola. Square term on y with linear x indicates a horizontal parabola.
Answer: e=ac where c=a2−b2. For ellipses, 0<e<1 since c<a always.
Answer: Hyperbola. Divide by 36 to get standard form with a2=9 and b2=4.
Answer: (x2−x1)2+(y2−y1)2. Derived from the Pythagorean theorem in coordinate geometry.
Answer: c2=a2+b2. For hyperbolas, c>a since the foci are outside the vertices.
Answer: x2=4py. Where p is the distance from vertex to focus and directrix.
Answer: y=−2. From x2=8y, we have 4p=8, so p=2 and directrix is y=−p.
Answer: y=ax2. Where a determines the width and opens vertically.
Answer: Ax2+Bxy+Cy2+Dx+Ey+F=0. The discriminant B2−4AC determines the conic type.
Answer: A set of points where the difference of distances to two foci is constant. The absolute value of the difference equals 2a, where a is the semi-major axis.
Answer: Hyperbola. Difference of squares form indicates a hyperbola with a2=b2=1.
Answer: (x−h)2+(y−k)2=r2. Standard form for any circle with center and radius specified.
Answer: Circle. Complete the square to verify equal coefficients for x2 and y2.
Answer: A fixed line used to define the parabola. Points on the parabola are equidistant from focus and directrix.
Answer: e=ac where c=a2−b2. For ellipses, 0<e<1 since c<a always.
Answer: A set of points equidistant from a fixed point (center). All points are the same distance from the center point.
Answer: y=±abx. The slopes are ±ab for the standard hyperbola form.
Answer: Hyperbola. Difference of squares with equal coefficients indicates a hyperbola.
Answer: A set of points equidistant from a point (focus) and a line (directrix). This property defines the parabola's unique geometric shape.
Answer: a in a2x2+b2y2=1. The larger denominator corresponds to the major axis length.
Answer: (x−h)2=4p(y−k). Vertex form for vertical parabolas with vertex at (h,k).
Answer: Circle. Complete the square to verify it has equal coefficients for squared terms.
Answer: Center: (0,0), a=2, b=3. Divide by 36: 4x2+9y2=1, so a2=4, b2=9.
Answer: A fixed line used to define the parabola. Points on the parabola are equidistant from focus and directrix.
Answer: a2x2+b2y2=1. Where a and b are the semi-major and semi-minor axis lengths.
Answer: y=±abx. Lines the hyperbola approaches as x and y approach infinity.
Answer: y=±abx. The slopes are ±ab for the standard hyperbola form.
Answer: x2+y2=r2. Where r is the radius from the center at (0,0).
Answer: x2+y2=r2. Where r is the radius from the center at (0,0).
Answer: Center: (0,0), a=2, b=3. Divide by 36: 4x2+9y2=1, so a2=4, b2=9.
Answer: e=ac. Where c is the focal distance and a is the semi-major axis.
Answer: c=a2−b2. For an ellipse, c<a since the foci are inside the ellipse.
Answer: (x−h)2+(y−k)2=r2. Standard form for any circle with center and radius specified.
Answer: (2, 5). The vertex form shows the vertex at (h,k) where the parabola turns.
Answer: Circle. Equal coefficients (25) for x2 and y2 terms indicate a circle.
Answer: x2=4py. Where p is the distance from vertex to focus and directrix.
Answer: e=ac. Where c is the focal distance and a is the semi-major axis.
Answer: y=±abx. Lines the hyperbola approaches as x and y approach infinity.
Answer: Hyperbola. Eccentricity greater than 1 distinguishes hyperbolas from other conics.
Answer: (x−h)2=4p(y−k). Vertex form for vertical parabolas with vertex at (h,k).
Answer: y=ax2. Where a determines the width and opens vertically.