Math 3 Quiz: Algebraic Constraints And Loci

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Algebraic Constraints And LociQuestion 1 of 15

The constraint $\frac{x^2}{16} + \frac{y^2}{9} = 1$ defines an ellipse. A point $(x, y)$ on this ellipse satisfies $y = \frac{3x}{4}$ . What is the sum of the $x$ -coordinates of all such points?

0

, because the line passes through the center of the ellipse at the origin

\frac{32}{5}

, because the intersection points have positive

x

-coordinates summing to this value

0

, because the two intersection points are symmetric about the

y

-axis

-\frac{32}{5}

, because one intersection has negative

x

-coordinate with larger absolute value

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Math 3 Quiz

Math 3 Quiz: Algebraic Constraints And Loci

Practice Algebraic Constraints And Loci in Math 3 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Algebraic Constraints And Loci, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 3.

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Question 1

The constraint $\frac{x^2}{16} + \frac{y^2}{9} = 1$ defines an ellipse. A point $(x, y)$ on this ellipse satisfies $y = \frac{3x}{4}$ . What is the sum of the $x$ -coordinates of all such points?

$0$ , because the line passes through the center of the ellipse at the origin
$\frac{32}{5}$ , because the intersection points have positive $x$ -coordinates summing to this value
$0$ , because the two intersection points are symmetric about the $y$ -axis (correct answer)
$-\frac{32}{5}$ , because one intersection has negative $x$ -coordinate with larger absolute value

Explanation: When you encounter problems involving lines intersecting conic sections like ellipses, you're looking for points that simultaneously satisfy both equations. The key insight is recognizing geometric relationships that can simplify your work. To find intersection points, substitute the line equation

y = \frac{3x}{4}

into the ellipse equation:

\frac{x^2}{16} + \frac{(\frac{3x}{4})^2}{9} = 1

Simplifying:

\frac{x^2}{16} + \frac{9x^2/16}{9} = 1

, which becomes

\frac{x^2}{16} + \frac{x^2}{16} = 1

, so

\frac{x^2}{8} = 1

. This gives us

x^2 = 8

, meaning

x = \pm 2\sqrt{2}

. The two intersection points are

(2\sqrt{2}, \frac{3\sqrt{2}}{2})

and

(-2\sqrt{2}, -\frac{3\sqrt{2}}{2})

. Their x-coordinates sum to

2\sqrt{2} + (-2\sqrt{2}) = 0

. Answer C is correct: the intersection points are symmetric about the y-axis because the line

y = \frac{3x}{4}

passes through the origin, which is the center of the ellipse. Answer A incorrectly suggests this symmetry is the reason the sum is zero, rather than an observation about the geometry. Answer B gives

\frac{32}{5}

, which results from calculation errors in the substitution process. Answer D provides the negative of answer B, showing the same computational mistake with an incorrect sign assumption. Remember: when a line through the origin intersects a conic section centered at the origin, the intersection points are always symmetric about the origin, making sums of coordinates equal zero.

Question 2

The constraint $xy = 6$ defines a locus in the coordinate plane. Which of the following points lies on the intersection of this locus with the circle $x^2 + y^2 = 13$ ?

$(2, 3)$ and $(-2, -3)$ only
$(3, 2)$ and $(-3, -2)$ only
$(2, 3)$ , $(-2, -3)$ , $(3, 2)$ , and $(-3, -2)$ (correct answer)
$(\sqrt{6}, \sqrt{6})$ and $(-\sqrt{6}, -\sqrt{6})$ only

Explanation: To find intersections, solve the system:

xy = 6

and

x^2 + y^2 = 13

. From the first equation,

y = \frac{6}{x}

. Substituting into the second:

x^2 + \frac{36}{x^2} = 13

. Multiplying by

x^2

x^4 + 36 = 13x^2

, so

x^4 - 13x^2 + 36 = 0

. Let

u = x^2

u^2 - 13u + 36 = 0

, which factors as

(u-4)(u-9) = 0

. So

x^2 = 4

x^2 = 9

, giving

x = \pm 2

x = \pm 3

. When

x = 2

y = 3

; when

x = -2

y = -3

; when

x = 3

y = 2

; when

x = -3

y = -2

. All four points satisfy both equations. Choices A and B are incomplete. Choice D gives points that don't satisfy

xy = 6

Question 3

The set of points $(x, y)$ satisfying $|x| + |y| = 4$ forms a square in the coordinate plane. If this locus is rotated $45°$ counterclockwise about the origin, the resulting locus can be described by which equation?

$\max(|x|, |y|) = 2\sqrt{2}$
$|x - y| + |x + y| = 4\sqrt{2}$ (correct answer)
$|x - y| + |x + y| = 8$
$x^2 + y^2 = 8$

Explanation: When you encounter problems involving geometric transformations of loci, think systematically about how rotations affect coordinate relationships. The original equation

|x| + |y| = 4

describes a square with vertices at

(4,0)

(0,4)

(-4,0)

, and

(0,-4)

. To find the equation after a

45°

counterclockwise rotation, use the rotation transformation:

x' = x\cos(45°) - y\sin(45°) = \frac{x-y}{\sqrt{2}}

and

y' = x\sin(45°) + y\cos(45°) = \frac{x+y}{\sqrt{2}}

. This means

x = \frac{x'+y'}{\sqrt{2}}

and

y = \frac{y'-x'}{\sqrt{2}}

. Substituting into the original equation:

\left|\frac{x'+y'}{\sqrt{2}}\right| + \left|\frac{y'-x'}{\sqrt{2}}\right| = 4

, which simplifies to

\frac{|x'+y'| + |y'-x'|}{\sqrt{2}} = 4

. Therefore,

|x'+y'| + |y'-x'| = 4\sqrt{2}

. Dropping the primes gives us

|x+y| + |x-y| = 4\sqrt{2}

, which is choice B. Choice A gives the equation of a square aligned with the axes, not rotated. Choice C has the right form but wrong coefficient—it uses 8 instead of

4\sqrt{2}

, likely from forgetting the

\sqrt{2}

factor in the rotation. Choice D describes a circle, not the rotated square. For transformation problems, always apply the transformation systematically to the coordinates, then substitute back into the original equation. Remember that rotating geometric figures often changes the form of their equations significantly.

Question 4

Consider the system of constraints: $x + y \leq 6$ , $x - y \geq -2$ , and $y \geq 1$ . The locus of points $(x, y)$ satisfying all three constraints simultaneously forms a triangular region. What are the coordinates of the vertices of this triangle?

$(1, 5)$ , $(5, 1)$ , and $(-1, 1)$
$(2, 4)$ , $(4, 2)$ , and $(0, 2)$
$(3, 3)$ , $(1, 1)$ , and $(-1, 3)$
$(5, 1)$ , $(-1, 1)$ , and $(2.5, 3.5)$ (correct answer)

Explanation: When you encounter a system of linear inequalities, you're looking for the region where all constraints overlap. The vertices of this feasible region occur where the boundary lines intersect. First, convert each inequality to its boundary line equation:

x + y = 6

x - y = -2

, and

y = 1

. To find the vertices, solve each pair of equations simultaneously. For

x + y = 6

and

y = 1

: Substituting gives

x + 1 = 6

, so

x = 5

. This yields vertex

(5, 1)

. For

x - y = -2

and

y = 1

: Substituting gives

x - 1 = -2

, so

x = -1

. This yields vertex

(-1, 1)

. For

x + y = 6

and

x - y = -2

: Adding these equations gives

2x = 4

, so

x = 2

. Substituting back:

2 + y = 6

, so

y = 4

. However, this point

(2, 4)

doesn't satisfy

y \geq 1

as a boundary—it's well within the feasible region. The actual third vertex occurs where one constraint becomes active. Since

y = 1

intersects the other two lines, and the feasible region is bounded by

x + y = 6

and

x - y = -2

above

y = 1

, the third vertex is

(2.5, 3.5)

, which lies on both non-horizontal boundaries. Answer choice A gives vertices that don't all satisfy the constraint

x - y \geq -2

. Choice B includes

(0, 2)

, which violates

x + y \leq 6

when checked with other constraints. Choice C contains points outside the actual feasible region. Always verify that each proposed vertex satisfies all original inequalities—this catches calculation errors and confirms your solution.

Question 5

A point $(x, y)$ satisfies the constraint $x^2 - 4x + y^2 \geq 0$ . The boundary of this region is a circle. What is the area of the region that does NOT satisfy this constraint?

$\pi$ square units
$2\pi$ square units
$4\pi$ square units (correct answer)
$16\pi$ square units

Explanation: Complete the square:

x^2 - 4x + y^2 \geq 0

becomes

(x-2)^2 - 4 + y^2 \geq 0

, or

(x-2)^2 + y^2 \geq 4

. This represents the exterior and boundary of a circle with center

(2, 0)

and radius

2

. The region that does NOT satisfy this constraint is the interior of the circle, which has area

\pi r^2 = \pi(2)^2 = 4\pi

. Choice A uses radius

1

instead of

2

. Choice B uses

2\pi r

(circumference formula) instead of

\pi r^2

. Choice D uses the diameter instead of radius in the area formula.

Question 6

The constraint $x^2 + 4y^2 - 2x + 16y + 13 = 0$ defines a conic section. After completing the square, what type of conic is this, and what is its eccentricity?

A parabola with eccentricity $e = 1$ and vertex at $(1, -2)$
An ellipse with eccentricity $e = \frac{\sqrt{3}}{2}$ and center at $(1, -2)$ (correct answer)
An ellipse with eccentricity $e = \frac{1}{2}$ and center at $(1, -2)$
A hyperbola with eccentricity $e = \frac{\sqrt{5}}{2}$ and center at $(1, -2)$

Explanation: When you encounter a general conic equation like this, your first step is completing the square to identify the conic type and key features. The presence of both

x^2

and

y^2

terms with different coefficients suggests an ellipse. Let's complete the square systematically. Starting with

x^2 + 4y^2 - 2x + 16y + 13 = 0

, group the x and y terms: For x terms:

x^2 - 2x = (x-1)^2 - 1

For y terms:

4y^2 + 16y = 4(y^2 + 4y) = 4[(y+2)^2 - 4] = 4(y+2)^2 - 16

Substituting back:

(x-1)^2 - 1 + 4(y+2)^2 - 16 + 13 = 0

Simplifying:

(x-1)^2 + 4(y+2)^2 = 4

Dividing by 4:

\frac{(x-1)^2}{4} + \frac{(y+2)^2}{1} = 1

This is an ellipse in standard form with center

(1, -2)

, where

a^2 = 4

and

b^2 = 1

, so

a = 2

and

b = 1

. The eccentricity is

e = \frac{c}{a}

where

c = \sqrt{a^2 - b^2} = \sqrt{4-1} = \sqrt{3}

. Therefore,

e = \frac{\sqrt{3}}{2}

. Choice A incorrectly identifies this as a parabola—parabolas have only one squared variable. Choice C has the wrong eccentricity calculation, likely confusing the relationship between a, b, and c. Choice D misidentifies the conic as a hyperbola, which would require subtraction between the squared terms. Study tip: Remember that ellipses have the form

\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1

with eccentricity

e = \frac{\sqrt{a^2-b^2}}{a}

when

a > b

Question 7

Consider the algebraic constraint $x^2 + y^2 - 6x + 2y + k = 0$ . For which value of $k$ does this constraint define a point locus rather than the empty set?

$k = 10$ , and the locus is a single point at $(3, -1)$ (correct answer)
$k = -10$ , and the locus is a circle with radius $\sqrt{10}$
$k = 10$ , and the locus is a circle with radius $\sqrt{10}$
$k = -10$ , and the locus is a single point at $(3, -1)$

Explanation: Completing the square:

(x^2 - 6x) + (y^2 + 2y) + k = 0

, so

(x-3)^2 - 9 + (y+1)^2 - 1 + k = 0

, giving

(x-3)^2 + (y+1)^2 = 10 - k

. For a point locus (not empty set), we need

10 - k \geq 0

, so

k \leq 10

. When

k = 10

, we get

(x-3)^2 + (y+1)^2 = 0

, which means

x = 3

and

y = -1

, giving a single point. When

k < 10

, we get a circle with positive radius. Choice A correctly identifies that

k = 10

gives a single point at

(3, -1)

. Choice B has the wrong sign for

k

and wrong interpretation. Choice C has the right

k

but wrong geometric interpretation. Choice D has wrong sign for

k

Question 8

The locus of points equidistant from the point $(0, 4)$ and the line $y = -2$ forms a parabola. What is the equation of the directrix of this parabola?

$x = 0$ , and the vertex of the parabola is at $(1, 0)$
$y = 4$ , and the vertex of the parabola is at $(0, 1)$
$y = -2$ , and the vertex of the parabola is at $(0, 1)$ (correct answer)
$y = -2$ , and the vertex of the parabola is at $(0, 4)$

Explanation: When you encounter a problem about the locus of points equidistant from a point and a line, you're dealing with the fundamental definition of a parabola. The given point is the focus, and the given line is the directrix. Let's identify what we have: the focus is at

(0, 4)

and the directrix is the line

y = -2

. The vertex of a parabola lies exactly halfway between the focus and directrix. Since the focus is at

y = 4

and the directrix is at

y = -2

, the vertex has a y-coordinate of

\frac{4 + (-2)}{2} = 1

. The x-coordinate stays the same as the focus since this parabola opens vertically, so the vertex is at

(0, 1)

. The directrix is simply the given line that, along with the focus, defines the parabola. Therefore, the directrix is

y = -2

. Choice A incorrectly identifies the directrix as

x = 0

(a vertical line) and places the vertex at

(1, 0)

. Choice B incorrectly claims the directrix is

y = 4

, which is actually the y-coordinate of the focus, not the directrix. It also misplaces the vertex. Choice D correctly identifies the directrix as

y = -2

but incorrectly states the vertex is at

(0, 4)

, which is actually the focus. Choice C correctly identifies both the directrix as

y = -2

and the vertex at

(0, 1)

. Remember: in parabola problems defined by focus and directrix, the directrix is the given line, and the vertex is always the midpoint between the focus and directrix.

Question 9

The constraint $|x - 2| + |y + 1| \leq 3$ defines a locus in the coordinate plane. What is the area of this region?

$18$ square units, forming a diamond shape with vertices at four cardinal points (correct answer)
$9\pi$ square units, forming a circle centered at $(2, -1)$ with radius $3$
$12$ square units, forming a square with side length $\sqrt{12}$ units
$18$ square units, forming a square with vertices at $(5, -1)$ , $(2, 2)$ , $(-1, -1)$ , and $(2, -4)$

Explanation: The constraint

|x - 2| + |y + 1| \leq 3

represents all points within distance 3 from

(2, -1)

using the taxicab (Manhattan) metric. This forms a diamond (rhombus) rotated 45° with vertices at

(2+3, -1) = (5, -1)

(2, -1+3) = (2, 2)

(2-3, -1) = (-1, -1)

, and

(2, -1-3) = (2, -4)

. The diamond has diagonals of length 6 (horizontal) and 6 (vertical), so its area is

\frac{1}{2} \times 6 \times 6 = 18

square units. Choice B incorrectly applies Euclidean distance. Choice C gives wrong area calculation. Choice D has the correct area and vertices but incorrectly describes the shape as a square rather than a diamond.

Question 10

The locus defined by the constraint $\sqrt{(x-1)^2 + (y-2)^2} + \sqrt{(x-5)^2 + (y-2)^2} = 6$ forms an ellipse. What is the length of the minor axis of this ellipse?

$2\sqrt{3}$ units, with the major axis having length $6$ units
$2\sqrt{5}$ units, with foci located at $(1, 2)$ and $(5, 2)$ (correct answer)
$4\sqrt{2}$ units, with the center of the ellipse at $(3, 2)$
$2\sqrt{5}$ units, with the major axis aligned vertically along $x = 3$

Explanation: When you see an equation involving the sum of distances from two fixed points, you're looking at the definition of an ellipse. The equation

\sqrt{(x-1)^2 + (y-2)^2} + \sqrt{(x-5)^2 + (y-2)^2} = 6

represents all points where the sum of distances to the foci at

(1,2)

and

(5,2)

equals 6. First, identify the key parameters. The foci are at

(1,2)

and

(5,2)

, so the center is at their midpoint:

(\frac{1+5}{2}, \frac{2+2}{2}) = (3,2)

. The distance between foci is

2c = 5-1 = 4

, so

c = 2

. Since the sum of distances equals 6, we have

2a = 6

, giving us

a = 3

. Using the relationship

c^2 = a^2 - b^2

for ellipses:

4 = 9 - b^2

, so

b^2 = 5

and

b = \sqrt{5}

. The minor axis length is

2b = 2\sqrt{5}

. Choice A gives the wrong minor axis length (

2\sqrt{3}

) and incorrectly states the major axis is 6 units when it's actually

2a = 6

. Choice C provides the wrong minor axis length (

4\sqrt{2}

) but correctly identifies the center. Choice D has the correct minor axis length but wrongly claims the major axis is vertical—since the foci share the same y-coordinate, the major axis is horizontal. Choice B correctly gives the minor axis length as

2\sqrt{5}

units and accurately identifies the foci locations. Remember: for ellipse problems, always identify the foci first, then use

2a

(sum of distances),

2c

(distance between foci), and

c^2 = a^2 - b^2

to find all dimensions.

Question 11

Consider the constraint $xy = 12$ . The points on this locus that are closest to the origin have coordinates satisfying which additional relationship?

$x^2 = y^2$ , and the minimum distance to the origin is $4\sqrt{3}$ units
$x = -y$ , and the minimum distance to the origin is $2\sqrt{6}$ units
$|x| = |y|$ , and the minimum distance to the origin is $4\sqrt{3}$ units
$x^2 + y^2 = 24$ , and the minimum distance to the origin is $2\sqrt{6}$ units (correct answer)

Explanation: We want to minimize

\sqrt{x^2 + y^2}

subject to

xy = 12

. Since

y = \frac{12}{x}

, we minimize

f(x) = x^2 + \frac{144}{x^2}

. Taking the derivative:

f'(x) = 2x - \frac{288}{x^3} = 0

, so

2x = \frac{288}{x^3}

, giving

2x^4 = 288

, thus

x^4 = 144

, and

x^2 = 12

. So

x = \pm 2\sqrt{3}

. When

x = 2\sqrt{3}

y = \frac{12}{2\sqrt{3}} = 2\sqrt{3}

. When

x = -2\sqrt{3}

y = \frac{12}{-2\sqrt{3}} = -2\sqrt{3}

. At these points,

x^2 + y^2 = 12 + 12 = 24

, and the distance to origin is

\sqrt{24} = 2\sqrt{6}

. Choice D correctly identifies both conditions. Choice A has

x^2 = y^2

which is true (both equal 12), but wrong distance. Choice B requires

x = -y

which only holds for one of the two closest points. Choice C has

|x| = |y|

which is correct, but wrong distance calculation.

Question 12

The locus of points equidistant from the lines $3x + 4y = 12$ and $3x + 4y = -8$ is:

The line $3x + 4y = 10$ only
The line $3x + 4y = 2$ only (correct answer)
Two parallel lines: $3x + 4y = 2$ and $4x + 3y = 0$
The line $4x - 3y = 0$ only

Explanation: When you encounter a question about finding points equidistant from two lines, you're looking for the locus that forms a perpendicular bisector between parallel lines, or the angle bisectors between intersecting lines. First, notice that both given lines have the form

3x + 4y = c

, meaning they're parallel since they have identical coefficients for

x

and

y

. For parallel lines, the locus of equidistant points is simply the line that runs exactly halfway between them. To find this midline, you need the average of the constant terms. The first line is

3x + 4y = 12

and the second is

3x + 4y = -8

. The midpoint value is

\frac{12 + (-8)}{2} = \frac{4}{2} = 2

. Therefore, the locus is

3x + 4y = 2

. Looking at the wrong answers: Choice A gives

3x + 4y = 10

, which would be the average of 12 and 8 (not -8) - a common sign error. Choice C incorrectly suggests two parallel lines exist, which only happens when you're finding points at a fixed distance (not equidistant) from a single line. Choice D gives

4x - 3y = 0

, which represents a perpendicular line to the original pair, but perpendicular doesn't mean equidistant. Remember: for parallel lines of the form

ax + by = c_1

and

ax + by = c_2

, the equidistant locus is always

ax + by = \frac{c_1 + c_2}{2}

. Watch your signs carefully when averaging the constants.

Question 13

The set of points $(x, y)$ satisfying $x^2 + y^2 - 4x + 6y \leq 12$ represents which of the following?

The exterior and boundary of a circle with center $(2, -3)$ and radius $5$
The interior and boundary of a circle with center $(-2, 3)$ and radius $5$
The interior and boundary of a circle with center $(2, -3)$ and radius $25$
The interior and boundary of a circle with center $(2, -3)$ and radius $5$ (correct answer)

Explanation: When you encounter an inequality involving

x^2

and

y^2

terms, you're looking at a circle problem. The key is converting the given form into standard circle form:

(x-h)^2 + (y-k)^2 = r^2

, where

(h,k)

is the center and

r

is the radius. Starting with

x^2 + y^2 - 4x + 6y \leq 12

, you need to complete the square for both variables. For the

x

terms:

x^2 - 4x

becomes

(x-2)^2 - 4

since

(-4/2)^2 = 4

. For the

y

terms:

y^2 + 6y

becomes

(y+3)^2 - 9

since

(6/2)^2 = 9

. Substituting back:

(x-2)^2 - 4 + (y+3)^2 - 9 \leq 12

, which simplifies to

(x-2)^2 + (y+3)^2 \leq 25

. This represents all points whose distance from

(2, -3)

is at most

5

(since

\sqrt{25} = 5

). The "less than or equal to" means you include both the interior and boundary of the circle. Choice A incorrectly states "exterior and boundary" - the

\leq

symbol means interior, not exterior. Choice B has the wrong center

(-2, 3)

instead of

(2, -3)

- this comes from sign errors when completing the square. Choice C correctly identifies the center but confuses

r^2 = 25

with

r = 25

, giving radius

25

instead of

5

. Remember: when completing the square, watch your signs carefully, and distinguish between

r^2

and

r

when finding the radius. The inequality symbol determines whether you're inside (

\leq

) or outside (

\geq

) the circle.

Question 14

The locus of points $(x, y)$ satisfying $|x - y| = 3$ consists of two lines. If a point on this locus has $x$ -coordinate $7$ , what are the possible $y$ -coordinates?

$y = 4$ and $y = 10$ (correct answer)
$y = -4$ and $y = 4$
$y = 3$ and $y = -3$
$y = 7$ and $y = -7$

Explanation: The equation

|x - y| = 3

means either

x - y = 3

x - y = -3

. This gives us

y = x - 3

y = x + 3

. When

x = 7

, we get

y = 7 - 3 = 4

y = 7 + 3 = 10

. Choice B uses

x = 0

instead of

x = 7

. Choice C incorrectly assumes

|y| = 3

. Choice D incorrectly assumes

|x| = |y|

Question 15

Consider the set of all points $(x, y)$ in the coordinate plane such that $\sqrt{(x-2)^2 + (y+1)^2} + \sqrt{(x+2)^2 + (y+1)^2} = 8$ . Which of the following best describes this locus?

An ellipse with center at $(0, -1)$ and vertices at $(-4, -1)$ and $(4, -1)$ (correct answer)
A hyperbola with center at $(0, -1)$ and vertices at $(-3, -1)$ and $(3, -1)$
An ellipse with center at $(0, -1)$ and vertices at $(-3, -1)$ and $(3, -1)$
A circle with center at $(0, -1)$ and radius $4$

Explanation: This equation represents the sum of distances from point

(x,y)

to the two foci

(-2, -1)

and

(2, -1)

equals 8. This is the definition of an ellipse with

2a = 8

, so

a = 4

. The distance between foci is

2c = 4

, so

c = 2

. The center is at the midpoint of the foci:

(0, -1)

. The vertices are at

(-4, -1)

and

(4, -1)

. Choice B incorrectly identifies it as a hyperbola. Choice C has the wrong vertices (uses

a = 3

instead of

a = 4

). Choice D incorrectly identifies it as a circle.