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What are the conic sections, and how do their standard equations encode their shape and key features?

Topic 4.6 Conic Sections: identify and analyze parabolas, ellipses, circles and hyperbolas from their equations, and describe their key features.

A focused answer to AP Precalculus Topic 4.6, covering the four conic sections, their standard implicit equations, how to read center, radius, vertices and orientation from the equation, and how to tell the conics apart.

Generated by Claude Opus 4.89 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this topic is asking
  2. The four standard forms
  3. Telling the conics apart
  4. The circle as a special ellipse
  5. Try this

What this topic is asking

The College Board (Topic 4.6) wants you to identify and analyze the conic sections: parabolas, ellipses, circles and hyperbolas. Each has a standard implicit equation whose structure encodes the shape, and from that equation you read the center, radius or semi-axes, vertices and orientation.

The four standard forms

The constants (h,k)(h, k) are the shifts that move the center or vertex away from the origin, exactly the transformations of Topic 1.11.

Telling the conics apart

This algebraic signature identifies the conic before you compute any feature, which makes the analysis systematic.

The circle as a special ellipse

A point worth stating once is that a circle is the special case of an ellipse with equal semi-axes. When a=ba = b the ellipse equation (xβˆ’h)2a2+(yβˆ’k)2a2=1\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{a^2} = 1 multiplies out to (xβˆ’h)2+(yβˆ’k)2=a2(x - h)^2 + (y - k)^2 = a^2, a circle of radius aa. So the family runs continuously: stretch a circle unequally and it becomes an ellipse; the circle is just the symmetric member. Recognizing this keeps you from treating the circle and ellipse as unrelated and explains why the circle's parametrisation (Topic 4.4) generalizes to the ellipse by using different radii in the cosine and sine terms.

A related point concerns the parabola, which stands apart from the other three. The circle, ellipse and hyperbola all have two squared terms, so they are symmetric about their center; the parabola has only one squared term and so has a single vertex rather than a center, and it opens in one direction only. In y=a(xβˆ’h)2+ky = a(x - h)^2 + k the vertex is (h,k)(h, k) and the sign of aa tells you whether it opens up (a>0a > 0) or down (a<0a < 0); the form x=a(yβˆ’k)2+hx = a(y - k)^2 + h opens right or left instead. This is the same vertex-form analysis you met for quadratic polynomials in Unit 1, now seen as one of the four conic sections. Keeping the parabola's single-squared-term signature in mind is the fastest way to separate it from the ellipse and hyperbola at a glance.

Try this

Q1. What conic is x2+y2=49x^2 + y^2 = 49, and what is its radius? [1 point]

  • Cue. A circle centered at the origin with radius 49=7\sqrt{49} = 7.

Q2. Which way does y24βˆ’x29=1\frac{y^2}{4} - \frac{x^2}{9} = 1 open? [1 point]

  • Cue. The yy-term is positive, so the hyperbola opens up and down.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2025 (style)1 marksSection I, Part A (multiple choice, no calculator). The equation x29+y24=1\frac{x^2}{9} + \frac{y^2}{4} = 1 represents which conic? (A) A circle (B) An ellipse (C) A parabola (D) A hyperbola
Show worked answer β†’

The correct answer is (B), an ellipse.

A sum of two squared terms set equal to 11, with different positive denominators, is an ellipse. Here the semi-axes are 9=3\sqrt{9} = 3 horizontally and 4=2\sqrt{4} = 2 vertically. If the denominators were equal it would be a circle; a difference of squares would be a hyperbola.

AP 2025 (style)4 marksSection II (free response, no calculator). Consider x216βˆ’y29=1\frac{x^2}{16} - \frac{y^2}{9} = 1. (a) Identify the conic and its center. (b) Find the vertices and state the orientation.
Show worked answer β†’

A 4-point question on analyzing a hyperbola.

(a) Conic and center (2 points): a difference of squared terms equal to 11 is a hyperbola, centered at the origin (0,0)(0, 0) since there are no shifts in xx or yy.
(b) Vertices and orientation (2 points): the positive term is the xx-term, so the hyperbola opens left-right. The vertices are at (Β±a,0)(\pm a, 0) with a=16=4a = \sqrt{16} = 4, that is (4,0)(4, 0) and (βˆ’4,0)(-4, 0). The transverse axis is horizontal.

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