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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: 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: 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: 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: 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: Ellipse. Divide by 144 to get standard form with positive coefficients.
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: a2x2−b2y2=1. Subtraction between squared terms indicates a hyperbola opening horizontally.
Answer: (4, 0). From y2=16x, we get 4p=16, so p=4 and focus is at (p,0).
Answer: y=−2. From x2=8y, we have 4p=8, so p=2 and directrix is y=−p.
Answer: e=ac where c=a2−b2. For ellipses, 0<e<1 since c<a always.
Answer: (x2−x1)2+(y2−y1)2. Derived from the Pythagorean theorem in coordinate geometry.
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: A fixed line used to define the parabola. Points on the parabola are equidistant from focus and directrix.
Answer: A set of points equidistant from a fixed point (center). All points are the same distance from the center point.
Answer: a in a2x2+b2y2=1. The larger denominator corresponds to the major axis length.
Answer: Circle. Complete the square to verify it has equal coefficients for squared terms.
Answer: a2x2+b2y2=1. Where a and b are the semi-major and semi-minor axis lengths.
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: 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: x2=4py. Where p is the distance from vertex to focus and directrix.
Answer: y=±abx. Lines the hyperbola approaches as x and y approach infinity.
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.