Conic Sections - AP Precalculus

A spacecraft's orbit is modeled by an ellipse, as in Kepler's Laws. The standard form is $\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1.$ Here, $(h,k)$ is the center, $a$ is the semi-major axis, and $b$ is the semi-minor axis. Engineers adjust $a$ to widen or tighten the orbit while keeping the center fixed. Refer to the equation provided in the passage: how does increasing $a$ affect the ellipse's horizontal extent?
Answer: It increases the horizontal radius (semi-major axis). This question tests AP Precalculus understanding of conic sections, particularly the ability to interpret and manipulate ellipse equations. In the standard form of an ellipse, $\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$ , the parameter $a$ represents the horizontal semi-axis length, determining how far the ellipse extends horizontally from its center. The passage describes how engineers adjust $a$ to modify the spacecraft's orbital width while maintaining the same center position. Choice A correctly identifies that increasing $a$ increases the horizontal radius because in the ellipse equation, a larger $a$ value in the denominator means the ellipse extends farther along the x-axis. Choice C is incorrect because it confuses the role of $a$ as a scaling parameter with the center coordinates $(h,k)$ , a common mistake when students misinterpret the standard form. To help students: Use visual demonstrations showing how changing $a$ stretches or compresses the ellipse horizontally while the center remains fixed. Watch for: confusion between parameters that affect size versus position in conic equations.
In acoustics, hyperbolas can model locations with equal sound intensity. Identify the conic section represented by $\frac{(x-2)^2}{9}-\frac{(y+1)^2}{4}=1$ based on the equation provided in the passage.
Answer: Hyperbola This question tests AP Precalculus understanding of conic sections, particularly the ability to identify different conic types from their equations. The key distinguishing feature is the operation between the two squared terms: addition creates circles or ellipses, while subtraction creates hyperbolas, and having only one squared term creates parabolas. In the passage, the equation $\frac{(x-2)^2}{9}-\frac{(y+1)^2}{4}=1$ shows a subtraction between the squared terms, which is the defining characteristic of a hyperbola. Choice C correctly identifies this as a hyperbola because of the minus sign between the fractions, which matches the standard form $\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$ . Choice A is incorrect because ellipses require addition between the terms, not subtraction, a common mistake when students focus on the fraction format without noticing the operation. To help students: Create a decision tree showing that subtraction always indicates hyperbolas regardless of other parameters. Practice identifying conic types by first looking for the operation (+, -, or single squared term) before analyzing other features.
Architects design circular domes using $ $(x-h)^2$ + $(y-k)^2$ = $r^2$ $. Based on the conic section described, which parameter change increases the circle's size without moving its center?
Answer: Increase $r$ This question tests AP Precalculus understanding of conic sections, particularly the ability to interpret circle equations and their parameters. In the standard form of a circle $(x-h)^2+(y-k)^2=r^2$ , the parameter $r$ represents the radius, $(h,k)$ represents the center, and changing $r$ affects only the size while changing $h$ or $k$ affects only the position. In the passage, the equation demonstrates that to increase a circle's size without moving its center, one must increase the radius parameter $r$ . Choice A correctly identifies that increasing $r$ makes the circle larger while keeping the center at $(h,k)$ unchanged, which directly follows from the geometric definition of a circle. Choice B is incorrect because increasing $h$ would shift the circle horizontally to the right, a common mistake when students confuse size changes with position changes. To help students: Use concentric circles to demonstrate how different radii create different sized circles with the same center. Emphasize that $(h,k)$ controls position while $r$ controls size, and these are independent properties.
Kepler's Law models an orbit by $\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$ . Refer to the equation provided in the passage: which equation represents an ellipse centered at $(1,-2)$ with $a=5$ and $b=3$ ?
Answer: $\frac{(x-1)^2}{25}+\frac{(y+2)^2}{9}=1$ This question tests AP Precalculus understanding of conic sections, particularly the ability to construct ellipse equations from given parameters. The standard form of an ellipse is $\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$ where $(h,k)$ is the center and $a$ , $b$ are the semi-axis lengths. In the passage, we need an ellipse centered at $(1,-2)$ with $a=5$ and $b=3$ , so we substitute: $h=1$ , $k=-2$ , $a^2=25$ , and $b^2=9$ . Choice A correctly shows $\frac{(x-1)^2}{25}+\frac{(y+2)^2}{9}=1$ , properly placing $(y-(-2))=(y+2)$ in the equation. Choice C is incorrect because it uses $a$ and $b$ instead of $a^2$ and $b^2$ in the denominators. To help students: Emphasize that denominators must be $a^2$ and $b^2$ , not just $a$ and $b$ , and practice converting between center coordinates and equation form. Watch for: forgetting to square the semi-axis lengths and sign errors with negative center coordinates.
Kepler's orbital model uses $\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$ . Based on the conic section described, identify the conic represented by $\frac{(x-2)^2}{9}+\frac{(y+1)^2}{4}=1$ .
Answer: Ellipse centered at $(2,-1)$ This question tests AP Precalculus understanding of conic sections, particularly the ability to identify conic types from their equations. The equation $\frac{(x-2)^2}{9}+\frac{(y+1)^2}{4}=1$ follows the standard form of an ellipse $\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$ where both terms are positive and added together. In the passage, we can identify that $h=2$ , $k=-1$ , $a^2=9$ , and $b^2=4$ , giving us center $(2,-1)$ . Choice A correctly identifies this as an ellipse centered at $(2,-1)$ because it has the characteristic form with two positive squared terms that sum to 1. Choice B is incorrect because hyperbolas have a minus sign between terms, not a plus sign. To help students: Create a flowchart for identifying conics - look for plus/minus signs, number of squared terms, and whether they equal 1 or 0. Watch for: students confusing the sign of $k$ when the equation shows $(y+1)^2$ rather than $(y-(-1))^2$ .
In projectile motion, a ball's path often follows a parabola described by $y=ax^2+bx+c.$ The constant $a$ controls how sharply the path curves, while $b$ and $c$ affect tilt and vertical placement. Coaches compare two throws by changing only $a$ and keeping $b$ and $c$ fixed. Based on the conic section described, what is the effect of increasing $|a|$ on the trajectory's shape?
Answer: The parabola becomes narrower and more curved. This question tests AP Precalculus understanding of conic sections, particularly the ability to interpret parabola equations and their parameters. In the quadratic form $y=ax^2+bx+c$ , the coefficient $a$ controls the parabola's curvature and opening direction, with larger $|a|$ values creating sharper curves. The passage explains how coaches compare throws by varying only the parameter $a$ while keeping other parameters constant. Choice B correctly identifies that increasing $|a|$ makes the parabola narrower and more curved because a larger absolute value of $a$ causes the y-values to change more rapidly as x changes, creating a steeper curve. Choice A is incorrect because it reverses the relationship between $|a|$ and curvature, a common misconception when students think larger values always mean wider shapes. To help students: Graph multiple parabolas with different $a$ values on the same axes to visualize the effect. Watch for: confusion about how the magnitude versus sign of $a$ affects the parabola's shape and direction.
An elliptical orbit is written in standard form as $\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1,$ where $(h,k)$ is the center and $a,b>0$ set the radii. In mission planning, changing $h$ and $k$ repositions the orbit without changing its size. Based on the passage, what is the effect of increasing $h$ while holding $a,b,$ and $k$ constant?
Answer: The ellipse shifts right with the same radii. This question tests AP Precalculus understanding of conic sections, particularly how center parameters affect ellipse position. In the standard ellipse equation $\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$ , the parameters $(h,k)$ determine the center location while $a$ and $b$ control the size. Increasing $h$ while keeping other parameters constant shifts the entire ellipse horizontally to the right without changing its shape or size. Choice A correctly identifies that the ellipse shifts right with the same radii because changing $h$ only affects the x-coordinate of the center, translating the entire figure horizontally. Choice B is incorrect because it confuses center translation with size change, mistaking the role of $h$ (position) with the role of $a$ (horizontal radius). To help students: Use transformation sequences showing how changing different parameters affects the ellipse independently. Watch for: mixing up which parameters control position versus size in conic equations.
Acoustics teams can locate a sound source using hyperbolas: points with a constant difference in distances to two microphones form a hyperbola. A standard form is $\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1,$ where $(h,k)$ is the center and $a,b>0$ scale the branches. Based on the conic section described, identify the conic represented by $\frac{(x+3)^2}{16}-\frac{(y-2)^2}{4}=1$ .
Answer: Hyperbola centered at $(-3,2)$ . This question tests AP Precalculus understanding of conic sections, particularly the ability to identify hyperbolas from their standard equations. The equation $\frac{(x+3)^2}{16}-\frac{(y-2)^2}{4}=1$ follows the standard hyperbola form $\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$ with a subtraction between the squared terms. The center is at $(h,k) = (-3,2)$ since $(x+3)^2 = (x-(-3))^2$ and $(y-2)^2$ appear in the equation. Choice B correctly identifies this as a hyperbola centered at $(-3,2)$ because the equation has the characteristic subtraction sign between squared terms that defines a hyperbola. Choice A is incorrect because an ellipse would have addition instead of subtraction between the squared terms, a fundamental distinction students must recognize. To help students: Emphasize that the operation (+ or -) between squared terms determines whether it's an ellipse/circle (+) or hyperbola (-). Watch for: students focusing only on the center coordinates while overlooking the critical sign difference.
In an orbital ellipse, $\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$ where $(h,k)$ is the center. Refer to the equation provided in the passage: which ordered pair is the center of $\frac{(x+3)^2}{16}+\frac{(y-5)^2}{25}=1$ ?
Answer: $(-3,5)$ This question tests AP Precalculus understanding of conic sections, particularly the ability to identify the center of an ellipse from its equation. In the standard form $\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$ , the center is at $(h,k)$ where the expressions are $(x-h)$ and $(y-k)$ . In the passage equation $\frac{(x+3)^2}{16}+\frac{(y-5)^2}{25}=1$ , we need to rewrite $(x+3)^2$ as $(x-(-3))^2$ to identify $h=-3$ , and $(y-5)^2$ already shows $k=5$ . Choice B correctly identifies the center as $(-3,5)$ by recognizing that $(x+3)=(x-(-3))$ . Choice A is incorrect because it fails to account for the sign change when $(x+3)$ is rewritten in standard form. To help students: Always rewrite equations in the form $(x-h)$ and $(y-k)$ before identifying the center, and practice with equations that have addition inside the parentheses. Watch for: sign errors when converting from $(x+a)$ to $(x-h)$ form.
Hyperbolas in acoustics use $\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$ . Based on the conic section described, which equation represents a hyperbola centered at $(0,0)$ ?
Answer: $\frac{x^2}{16}-\frac{y^2}{9}=1$ This question tests AP Precalculus understanding of conic sections, particularly the ability to identify hyperbolas from their equations and recognize standard forms. A hyperbola in standard form has the structure $\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$ or its vertical transverse axis variant, with the key feature being subtraction between the squared terms. In the passage, the standard form shows that a hyperbola centered at $(0,0)$ would have $h=0$ and $k=0$ , simplifying to $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ . Choice D correctly represents a hyperbola centered at $(0,0)$ with the equation $\frac{x^2}{16}-\frac{y^2}{9}=1$ , showing the characteristic subtraction between squared terms. Choice A is incorrect because it uses addition instead of subtraction, making it an ellipse rather than a hyperbola, a critical distinction students must recognize. To help students: Emphasize that the operation between terms (+ for ellipse/circle, - for hyperbola) is the primary identifier. Create a reference chart showing all conic sections in standard form to highlight their structural differences.
A dome's cross-section can be modeled by a circle $ $(x-h)^2$ + $(y-k)^2$ = $r^2$ $. Based on the conic section described, which equation represents a circle centered at $(4,-2)$ with radius $7$ ?
Answer: $(x-4)^2+(y+2)^2=49$ This question tests AP Precalculus understanding of conic sections, particularly the ability to write circle equations given center and radius. The standard form $(x-h)^2+(y-k)^2=r^2$ requires substituting the center $(h,k)=(4,-2)$ and radius $r=7$ , being careful with signs in the binomials. In the passage, the standard form shows that for center $(4,-2)$ , we need $(x-4)^2$ and $(y-(-2))^2=(y+2)^2$ , with $r^2=49$ . Choice A correctly writes the equation as $(x-4)^2+(y+2)^2=49$ , properly handling the negative $y$ -coordinate by writing $(y-(-2))^2=(y+2)^2$ . Choice C is incorrect because it uses $r=7$ instead of $r^2=49$ on the right side, a common mistake when students forget to square the radius. To help students: Emphasize that the right side must be $r^2$ , not $r$ . Practice converting between different representations, always checking that $(y-(-2))=(y+2)$ for negative coordinates.
Elliptical orbits use $\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$ . Refer to the equation provided in the passage: which equation has center $(-4,0)$ and $a^2=36$ , $b^2=16$ ?
Answer: $\frac{(x+4)^2}{36}+\frac{y^2}{16}=1$ This question tests AP Precalculus understanding of conic sections, particularly the ability to construct ellipse equations from given information. We need an ellipse with center $(-4,0)$ , so $h=-4$ and $k=0$ , with $a^2=36$ and $b^2=16$ as the squared semi-axis lengths. In the passage format $\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$ , we substitute to get $\frac{(x-(-4))^2}{36}+\frac{(y-0)^2}{16}=1$ , which simplifies to $\frac{(x+4)^2}{36}+\frac{y^2}{16}=1$ . Choice A correctly shows this equation with $(x+4)^2$ representing $(x-(-4))^2$ and the given values for $a^2$ and $b^2$ . Choice C is incorrect because it uses $a=6$ and $b=4$ instead of $a^2=36$ and $b^2=16$ in the denominators. To help students: Always work with the squared form of semi-axis lengths in the standard equation, and practice converting between different representations of the center. Watch for: using $a$ and $b$ instead of $a^2$ and $b^2$ , and sign errors with negative center coordinates.
In astronomy, Kepler described planetary orbits as ellipses. Consider the standard form $\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$ with $a>b>0$ , where $(h,k)$ is the center. Based on the conic section described, how does increasing $a$ affect the ellipse's horizontal extent?
Answer: It widens left and right, increasing the major axis length This question tests AP Precalculus understanding of conic sections, particularly the ability to interpret and manipulate ellipse equations. In the standard form of an ellipse $\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$ where $a>b>0$ , the parameter $a$ represents the semi-major axis length along the horizontal direction, while $b$ represents the semi-minor axis length along the vertical direction. In the passage, the equation demonstrates that increasing parameter $a$ directly increases the horizontal extent of the ellipse, making it wider from left to right. Choice A correctly identifies that increasing $a$ widens the ellipse left and right, increasing the major axis length to $2a$ , which aligns with the mathematical definition of an ellipse's semi-major axis. Choice B is incorrect because changing $a$ affects the shape, not the center position $(h,k)$ , a common mistake when students confuse shape parameters with translation parameters. To help students: Use visual demonstrations showing how changing $a$ stretches the ellipse horizontally while keeping the center fixed. Encourage students to graph ellipses with different $a$ values using technology to see the direct relationship between $a$ and horizontal width.
A dome's cross-section is modeled by a circle in standard form $(x-h)^2+(y-k)^2=r^2.$ Builders interpret $(h,k)$ as the center point on a coordinate grid and $r$ as the dome's radius. Refer to the equation provided in the passage: which equation represents a circle centered at $(3,-2)$ with radius $5$ ?
Answer: $(x-3)^2+(y+2)^2=25$ This question tests AP Precalculus understanding of conic sections, particularly the ability to write circle equations from given parameters. A circle with center $(3,-2)$ and radius $5$ follows the standard form $(x-h)^2+(y-k)^2=r^2$ where $(h,k)=(3,-2)$ and $r=5$ . Substituting these values gives $(x-3)^2+(y-(-2))^2=5^2$ , which simplifies to $(x-3)^2+(y+2)^2=25$ . Choice A correctly represents this circle because it properly substitutes the center coordinates with correct signs and squares the radius. Choice C is incorrect because it uses $r=5$ instead of $r^2=25$ on the right side, a common error when students forget to square the radius in the standard form. To help students: Practice converting between verbal descriptions and algebraic forms, emphasizing sign conventions for center coordinates. Watch for: errors in handling negative coordinates and forgetting to square the radius value.
For a projectile, the path is often modeled by $y=ax^2+bx+c.$ The constant $a$ controls opening direction and curvature: $a>0$ opens upward, and $a<0$ opens downward. Refer to the equation provided in the passage: based on the passage, what occurs when $a$ changes from $0.5$ to $-0.5$ ?
Answer: The parabola opens downward instead of upward. This question tests AP Precalculus understanding of conic sections, particularly how the sign of the leading coefficient affects parabola orientation. In the equation $y=ax^2+bx+c$ , positive $a$ creates an upward-opening parabola while negative $a$ creates a downward-opening parabola. When $a$ changes from $0.5$ to $-0.5$ , the sign change reverses the parabola's opening direction while the magnitude remains the same. Choice B correctly identifies that the parabola opens downward instead of upward because the negative coefficient causes the parabola to flip vertically. Choice A is incorrect because it reverses the relationship, suggesting the parabola was originally opening downward, which contradicts the positive initial value of $a=0.5$ . To help students: Use projectile motion examples where positive $a$ represents gravity-free motion and negative $a$ represents downward acceleration due to gravity. Watch for: confusion about which sign corresponds to which opening direction.
In astronomy, planetary paths can be approximated by ellipses following Kepler's Laws. A common model is $\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1,$ where $(h,k)$ is the center and $a,b>0$ set the horizontal and vertical radii. Identify the conic section represented by the equation $\frac{(x-2)^2}{25}+\frac{(y+1)^2}{9}=1$ .
Answer: Ellipse centered at $(2,-1)$ . This question tests AP Precalculus understanding of conic sections, particularly the ability to identify conic types from their standard equations. The given equation $\frac{(x-2)^2}{25}+\frac{(y+1)^2}{9}=1$ follows the standard ellipse form $\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$ with both squared terms having the same sign (positive) and being added. The center is at $(h,k) = (2,-1)$ since we have $(x-2)^2$ and $(y-(-1))^2 = (y+1)^2$ in the numerators. Choice A correctly identifies this as an ellipse centered at $(2,-1)$ because the equation matches the ellipse standard form with $a^2=25$ and $b^2=9$ . Choice D is incorrect because a circle would require $a^2=b^2$ , but here $25 \neq 9$ , making this an ellipse rather than a circle. To help students: Create a flowchart for identifying conic sections based on equation structure and coefficient relationships. Watch for: confusion between ellipses and circles, which are special cases of ellipses with equal radii.
Architects design circular domes using the circle equation $(x-h)^2+(y-k)^2=r^2.$ The center is $(h,k)$ and $r$ is the radius, which determines the dome's span. Suppose the design keeps $(h,k)$ fixed but changes $r$ . Refer to the equation provided in the passage: based on the passage, what is the effect of increasing $r$ ?
Answer: The circle enlarges while keeping the same center. This question tests AP Precalculus understanding of conic sections, particularly the ability to interpret circle equations and their parameters. In the standard circle equation $(x-h)^2+(y-k)^2=r^2$ , the parameter $r$ represents the radius, determining the circle's size while $(h,k)$ fixes its center position. The passage describes how architects modify the dome's span by changing $r$ while keeping the center coordinates constant. Choice A correctly identifies that increasing $r$ enlarges the circle while maintaining the same center because $r$ only affects the distance from center to circumference, not the center's location. Choice B is incorrect because it confuses the radius parameter with vertical translation, a common error when students mix up size and position parameters in conic equations. To help students: Use dynamic geometry software to show how changing $r$ affects only the circle's size. Watch for: misconceptions about which parameters control size versus position in standard form equations.
Circular arches in architecture use $(x-h)^2+(y-k)^2=r^2$ to encode center and radius. Students often confuse the sign inside parentheses when identifying the center. Refer to the equation provided in the passage: identify the circle's center for $(x+4)^2+(y-7)^2=36$ .
Answer: $(-4,7)$ This question tests AP Precalculus understanding of conic sections, particularly identifying circle centers from standard form equations. In the equation $(x+4)^2+(y-7)^2=36$ , students must recognize that $(x+4)^2 = (x-(-4))^2$ , revealing that $h=-4$ , while $(y-7)^2$ shows $k=7$ . The center is therefore at $(-4,7)$ , requiring careful attention to sign conventions. Choice B correctly identifies the center as $(-4,7)$ because the equation $(x+4)^2$ indicates the x-coordinate is -4 (opposite sign of what appears in the parentheses). Choice A is incorrect because it uses the wrong sign for the x-coordinate, taking +4 directly from the equation instead of recognizing the standard form requires $(x-h)^2$ . To help students: Practice rewriting equations like $(x+4)^2$ as $(x-(-4))^2$ to reinforce sign patterns. Watch for: the common error of using the sign that appears in the equation rather than the opposite sign required by standard form.
A projectile's path is modeled by $y=ax^2+bx+c$ . Based on the conic section described, which statement correctly links $a$ to the parabola's orientation?
Answer: The parabola opens upward when $a>0$ This question tests AP Precalculus understanding of conic sections, particularly the ability to connect parabola parameters to their geometric properties. In the quadratic form $y=ax^2+bx+c$ , the sign of the leading coefficient $a$ determines whether the parabola opens upward (when $a>0$ ) or downward (when $a<0$ ), which is a fundamental property of parabolas. In the passage, the equation form shows that the coefficient of $x^2$ directly controls the parabola's orientation, with positive values creating upward-opening curves. Choice A correctly states that the parabola opens upward when $a>0$ , which aligns with the standard behavior of quadratic functions. Choice B is incorrect because it reverses the relationship, claiming upward opening occurs when $a<0$ , a fundamental misunderstanding of how the leading coefficient affects parabola orientation. To help students: Use the analogy of positive $a$ creating a "smile" shape and negative $a$ creating a "frown" shape. Graph multiple parabolas with different signs of $a$ to reinforce this critical relationship.
In sound localization, a hyperbola can be modeled by $\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1.$ The parameter $a$ influences how far the vertices sit from the center along the transverse axis, affecting how "open" the branches appear. Refer to the equation provided in the passage: how does increasing $a$ affect the hyperbola's vertices?
Answer: They move farther from the center along the $x$ -direction. This question tests AP Precalculus understanding of conic sections, particularly how parameters affect hyperbola structure. In the standard hyperbola equation $\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$ , the parameter $a$ determines the distance from the center to the vertices along the transverse (horizontal) axis. The vertices are located at $(h±a, k)$ , so increasing $a$ moves them farther from the center horizontally. Choice A correctly identifies that vertices move farther from the center along the x-direction because larger $a$ values increase the distance between the center and vertices. Choice B is incorrect because it suggests vertices move closer, contradicting the direct relationship between $a$ and vertex distance, a common error when students confuse inverse and direct relationships. To help students: Draw hyperbolas with different $a$ values, marking vertex positions to visualize the relationship. Watch for: confusion about which axis is transverse and how parameters relate to geometric features.

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