Algebra 3 Quiz: Conics In General Form

20 questions · exam conditions

0:00

Conics In General FormQuestion 1 of 20

After completing the square, which standard form represents $9y^2-4x^2+18y+16x-43=0$ ?

\frac{(x-2)^2}{9}-\frac{(y+1)^2}{4}=1

\frac{(y-1)^2}{4}-\frac{(x+2)^2}{9}=1

\frac{(y+1)^2}{4}-\frac{(x-2)^2}{9}=1

\frac{(y+1)^2}{9}-\frac{(x-2)^2}{4}=1

← Back to quizzes

Algebra 3 Quiz

Algebra 3 Quiz: Conics In General Form

Practice Conics In General Form in Algebra 3 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Conics In General Form, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra 3.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

After completing the square, which standard form represents $9y^2-4x^2+18y+16x-43=0$ ?

$\frac{(x-2)^2}{9}-\frac{(y+1)^2}{4}=1$
$\frac{(y-1)^2}{4}-\frac{(x+2)^2}{9}=1$
$\frac{(y+1)^2}{4}-\frac{(x-2)^2}{9}=1$ (correct answer)
$\frac{(y+1)^2}{9}-\frac{(x-2)^2}{4}=1$

Explanation: Whenever you see a conic in expanded form, your first thought should be: complete the square for each variable to see which standard shape hides inside. Here, the equation has both

y^2

and

x^2

with opposite signs, so it is a hyperbola. Group the

y

-terms and

x

-terms separately. For

y

9y^2+18y=9(y^2+2y)=9[(y+1)^2-1]=9(y+1)^2-9

. For

x

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

. Substitute back:

9(y+1)^2-9-4(x-2)^2+16-43=0

. Simplify the constants:

-9+16-43=-36

, so

9(y+1)^2-4(x-2)^2-36=0

. Move the constant:

9(y+1)^2-4(x-2)^2=36

. Divide every term by

36

\frac{(y+1)^2}{4}-\frac{(x-2)^2}{9}=1

. That is the correct standard form. The choice with

\frac{(x-2)^2}{9}-\frac{(y+1)^2}{4}=1

reverses which term is positive, flipping the hyperbola's orientation. The choice with

(y-1)^2

and

(x+2)^2

uses the wrong signs inside the binomials, likely from misreading

y+1

y-1

x-2

x+2

. The choice with

\frac{(y+1)^2}{9}-\frac{(x-2)^2}{4}=1

swaps the denominators: after dividing by

36

, the

y

-coefficient

9

gives

1/4

, not

1/9

. Your strategy: always complete the square separately for each variable, factor out the leading coefficient first, then divide by the constant to get

1

. Double-check the binomial signs and denominators—they are the most common traps.

Question 2

The general equation $4x^2+9y^2-16x+18y-11=0$ is rewritten by completing the square. Which standard form is correct?

$\frac{(x-2)^2}{9}+\frac{(y+1)^2}{4}=1$ (correct answer)
$\frac{(x-2)^2}{4}+\frac{(y+1)^2}{9}=1$
$\frac{(x+2)^2}{9}+\frac{(y-1)^2}{4}=1$
$\frac{(x-2)^2}{9}+\frac{(y-1)^2}{4}=1$

Explanation: Whenever you see a general second-degree equation with positive

x^2

and

y^2

terms having different coefficients, think "ellipse." Your first move should be to group the

x

-terms and

y

-terms, then complete the square in each variable. Start with

4x^2-16x+9y^2+18y-11=0.

Factor the leading coefficients:

4(x^2-4x)+9(y^2+2y)-11=0.

Complete the squares:

4[(x-2)^2-4]+9[(y+1)^2-1]-11=0.

Distribute and simplify:

4(x-2)^2-16+9(y+1)^2-9-11=0

4(x-2)^2+9(y+1)^2=36.

Divide by 36:

\frac{(x-2)^2}{9}+\frac{(y+1)^2}{4}=1.

So the correct form has

x-2

and

y+1

, with the 9 under the

x

-term and 4 under the

y

-term. The choice with denominators swapped,

\frac{(x-2)^2}{4}+\frac{(y+1)^2}{9}=1

, treats the coefficients as if they belonged to the opposite variables. The choice with

\frac{(x+2)^2}{9}+\frac{(y-1)^2}{4}=1

reverses both center signs; completing the square gives

x-2

and

y+1

, not

x+2

and

y-1

. Finally,

\frac{(x-2)^2}{9}+\frac{(y-1)^2}{4}=1

has the correct

x

-part but the wrong

y

-sign: the original

+18y

term means the square is

(y+1)^2

, not

(y-1)^2

. Study tip: after completing the square, always check the signs inside the parentheses against the original linear terms, and check which denominator came from which coefficient.

Question 3

What is the graph in the real coordinate plane of $x^2+y^2-6x+4y+14=0$ ?

No real points in the plane (the empty set) (correct answer)
A circle centered at $(3,-2)$ with radius 1
A circle centered at $(3,-2)$ with radius $\sqrt{14}$
A single point located at $(3,-2)$

Explanation: Whenever you see an equation with

x^2

and

y^2

having equal coefficients, the graph is a circle, a point, or nothing at all. Your job is to complete the square on both variables to find the standard form. Group the

x

-terms and

y

-terms:

(x^2-6x)+(y^2+4y)+14=0.

Completing the squares gives

(x-3)^2-9+(y+2)^2-4+14=0,

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

Since a sum of two squares cannot be negative, there are no real points satisfying the equation. The graph is the empty set. The distractor "circle centered at

(3,-2)

with radius 1" would require the right side to be

1

, but here it is

-1

. The distractor"circle centered at

(3,-2)

with radius

\sqrt{14}

" mistakenly treats the original constant

14

as the squared radius, but completing the square shows the right side is actually negative. The distractor"single point located at

(3,-2)

" would require

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

which would happen if the constant were

13

instead of

14

; with

14

, the equation collapses to no real solution. Study tip: after completing the square, always check the sign of the number on the right. If it is positive, you have a circle with radius

\sqrt{\text{that number}}

; if zero, a single point; if negative, no real points. Don't mistake the original constant term fora radius—complete the square first.

Question 4

The equation $y^2+4y-2x+10=0$ represents which conic?

Parabola with vertex $(3,-2)$ opening left
Parabola with vertex $(3,-2)$ opening right (correct answer)
Parabola with vertex $(-3,2)$ opening right
Parabola with vertex $(-3,-2)$ opening right

Explanation: When you see an equation containing

y^2

but no

x^2

, you're dealing with a parabola that opens left or right. The key is to solve for

x

in terms of

y

, then complete the square to reveal the vertex. Starting with

y^2+4y-2x+10=0

, isolate

x

2x = y^2+4y+10

Complete the square on

y

y^2+4y+10 = (y+2)^2 + 6

2x = (y+2)^2 + 6

x = \frac12(y+2)^2 + 3

This is the form

x-h = a(y-k)^2

, where

h=3

k=-2

, and

a=\frac12>0

. Because

a

is positive, the parabola opens to the right, and the vertex is

(3,-2)

. The choice "vertex

(3,-2)

opening left" confuses the sign of

a

: a negative coefficient would open left, but here it's positive. "Vertex

(-3,2)

opening right" switches both coordinates, likely from misreading the completed-square signs. "Vertex

(-3,-2)

opening right" gets the

y

-coordinate right but flips the

x

-coordinate; remember

x-3

means the vertex's

x

-coordinate is

3

, not

-3

. A quick study tip: for

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

, the vertex is

(h,k)

, and if

a>0

it opens right; if

a<0

, it opens left. Isolate

x

, complete the square, and read the answer directly.

Question 5

Which general form is equivalent to $(x+2)^2+(y-5)^2=9$ ?

$x^2+y^2+4x-10y+20=0$ (correct answer)
$x^2+y^2+4x-10y+29=0$
$x^2+y^2+4x-10y+9=0$
$x^2+y^2+4x-10y+16=0$

Explanation: When you see an equation like this, recognize it as a circle in center-radius form. To convert to general form, expand each square, combine like terms, and set the equation equal to zero. Start with

(x+2)^2+(y-5)^2=9

. Expanding gives

x^2+4x+4+y^2-10y+25=9

. Combine the constant terms:

4+25=29

, so you have

x^2+y^2+4x-10y+29=9

. Then subtract

9

from both sides to get

x^2+y^2+4x-10y+20=0

, which is the correct equivalent form. Now look at the wrong choices. The version

x^2+y^2+4x-10y+29=0

is the trap of stopping before subtracting the

9

from both sides; it equals

9

, not

0

. The version

x^2+y^2+4x-10y+9=0

treats the radius squared

9

as the constant, but the constant must combine

4+25-9

, not simply reuse

9

. The version

x^2+y^2+4x-10y+16=0

likely comes from doing

25-9=16

while forgetting the

+4

from expanding

(x+2)^2

. Study tip: after expanding, always collect the constants before moving the right-hand side. Completing the square can verify your general form by recovering the original center and radius.

Question 6

What is the graph in the real coordinate plane of $x^2-y^2-4x+6y-5=0$ ?

A hyperbola centered at $(2,3)$
A parabola with vertex $(2,3)$
An ellipse centered at $(2,3)$
Two intersecting lines crossing at $(2,3)$ (correct answer)

Explanation: Whenever you see a conic equation with both

x^2

and

y^2

terms, completing the square in each variable should be your first move. Here,

x^2-4x=(x-2)^2-4

, and

-y^2+6y=-(y-3)^2+9

. Substituting into

x^2-y^2-4x+6y-5=0

gives

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

. This factors beautifully:

[(x-2)-(y-3)][(x-2)+(y-3)]=0

x-y+1=0

and

x+y-5=0

. So the graph is the union of two lines. Solving them together gives

x=2,\ y=3

, their intersection. Thus the graph is two intersecting lines crossing at

(2,3)

. Why not the other choices? A hyperbola centered at

(2,3)

would appear if the completed equation were

(x-2)^2-(y-3)^2=\text{positive constant}

. The opposite signs on the squared terms do suggest a hyperbola, but here the right-hand side is

0

, and the equation degenerates into a pair of lines. The parabola with vertex

(2,3)

would require something like

(y-3)^2=4(x-2)

, not a difference of squares. The ellipse centered at

(2,3)

would require both squared terms to have the same positive sign, not opposite signs. So those all miss the key feature:

A^2-B^2=0

factors into two lines. Study takeaway: when completing the square in a conic equation, always check the constant on the right-hand side. If you see

A^2-B^2=0

, factor immediately—the graph is two intersecting lines. If that constant is nonzero, then it is a hyperbola.

Question 7

What is the graph in the real coordinate plane of $x^2+y^2-6x+4y+13=0$ ?

A circle centered at $(3,-2)$ with radius 1
A circle centered at $(-3,2)$ with radius 1
A single point located at the coordinate $(3,-2)$ (correct answer)
No real points in the plane (the empty set)

Explanation: Whenever you see a quadratic in both

x

and

y

with equal squared coefficients, complete the square in

x

and

y

separately. That tells you exactly what shape the equation represents. . Group:

(x^2-6x)+(y^2+4y)+13=0

. Completing squares,

x^2-6x=(x-3)^2-9

and

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

. Substituting gives

(x-3)^2+(y+2)^2-9-4+13=0

,so

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

. Since a sum of squares equals zero only when each square is zero,

x=3

and

y=-2

. The graph is exactly the single point

(3,-2)

. The choice saying "a circle centered at

(3,-2)

with radius 1" treats the right side as

1

, but it is actually

0

; the radius would be

\sqrt{0}=0

. The choice saying "a circle centered at

(-3,2)

with radius 1" makes sign errors in the center and invents a radius: it would correspond to

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

. The choice saying "no real points" is tempting because a sum of squares might seem impossible, but it equals

0

at exactly one real point, so the graph is not empty. Strategy: after completing the square, compare the right side to zero: positive gives a circle, zero gives a single point, negative gives no real points. This pattern turns a tricky conic question into a quick check.

Question 8

For what value of $F$ does $x^2+y^2+8x-6y+F=0$ represent a circle with radius 3?

$F=9$
$F=25$
$F=34$
$F=16$ (correct answer)

Explanation: When you see a second-degree equation in

x

and

y

, the key is to recognize that a circle emerges from completing the square. For

x^2+y^2+8x-6y+F=0

, group the

x

terms and

y

terms:

x^2+8x

becomes

(x+4)^2-16

, and

y^2-6y

becomes

(y-3)^2-9

. Substituting back gives

(x+4)^2+(y-3)^2-25+F=0

, or

(x+4)^2+(y-3)^2=25-F

. The right side is

r^2

. You want radius 3, so

r^2=9

, meaning

25-F=9

, hence

F=16

. Why do the other values fail?

F=9

would make

r^2=16

, giving a radius of 4, not 3 — a common mistake is treating

F

as if it were already

r^2

F=25

makes

r^2=0

, which is a single point rather than a circle; this happens if you forget that the center's squared coordinates contribute to the constant term.

F=34

makes

r^2=-9

, impossible for a real circle; this is the trap of adding the radius squared

(16+9+9)

instead of subtracting it. The correct value,

F=16

, is exactly the constant needed to balance the completed-square form with radius 3. A reliable strategy: always complete the square first, then set the right side equal to

r^2

. Derive the constant by balancing, rather than memorizing a formula, to avoid sign errors. Quickly check your answer by plugging

F

back in and confirming the radius is indeed 3.

Question 9

The equation $3x^2-12x+2y+16=0$ represents which conic?

Parabola with vertex $(2,2)$ opening downward
Parabola with vertex $(-2,-2)$ opening upward
Parabola with vertex $(2,-2)$ opening downward (correct answer)
Parabola with vertex $(2,-2)$ opening upward

Explanation: Whenever you see a conic equation with only one squared variable, think parabola: the squared variable tells you the axis. Here

x

is squared and

y

is not, so it is a vertical parabola, and the sign of the

x^2

coefficient tells you which way it opens. Solve for

y

to put the equation in vertex form. From

3x^2-12x+2y+16=0

isolate

2y=-3x^2+12x-16

y=-\frac32 x^2+6x-8.

Complete the square on

x

y=-\frac32(x^2-4x)-8 = -\frac32\big((x-2)^2-4\big)-8 = -\frac32(x-2)^2+6-8 = -\frac32(x-2)^2-2.

In the form

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

, the vertex is

(h,k)=(2,-2)

, and since

a=-\frac32<0

, the parabola opens downward. The choice with vertex

(2,2)

comes from accidentally keeping the

+6

from completing the square but forgetting to subtract the original

-8

, shifting the

y

-coordinate up. The choice with vertex

(-2,-2)

opening upward reverses both the

x

-sign and the direction, as if using

(x+2)^2

with a positive coefficient. The choice with vertex

(2,-2)

opening upward has the correct vertex but ignores that the coefficient

a

is negative, so the opening cannot be upward. Study tip: for a vertical parabola, always rewrite as

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

. Read the vertex from

(h,k)

, then check the sign of

a

before choosing the opening direction.

Question 10

After completing the square, which standard form represents $x^2-4y^2+6x+8y-7=0$ ?

$\frac{(x+3)^2}{3}-\frac{(y-1)^2}{12}=1$
$\frac{(x+3)^2}{12}-\frac{(y-1)^2}{3}=1$ (correct answer)
$\frac{(y-1)^2}{3}-\frac{(x+3)^2}{12}=1$
$\frac{(x-3)^2}{12}-\frac{(y+1)^2}{3}=1$

Explanation: When you see a mixed quadratic equation like this, your first thought should be: conic section, and I need to complete the square separately for

x

and

y

. Here the

x^2

and

-4y^2

terms tell you it's a hyperbola. Group the

x

terms and

y

terms. For

x^2+6x

, complete the square to get

(x+3)^2-9

. For

-4y^2+8y

, factor out

-4

-4(y^2-2y)

, which becomes

-4[(y-1)^2-1]

, or

-4(y-1)^2+4

. Substitute back into the equation:

(x+3)^2-9-4(y-1)^2+4-7=0

Simplify:

(x+3)^2-4(y-1)^2-12=0

(x+3)^2-4(y-1)^2=12

Divide every term by 12:

\frac{(x+3)^2}{12}-\frac{(y-1)^2}{3}=1

So the correct standard form is

\frac{(x+3)^2}{12}-\frac{(y-1)^2}{3}=1

. The choice

\frac{(x+3)^2}{3}-\frac{(y-1)^2}{12}=1

reverses the denominators; that would mean the hyperbola opens vertically, not horizontally. The choice

\frac{(y-1)^2}{3}-\frac{(x+3)^2}{12}=1

also swaps the roles of

x

and

y

, again giving the wrong orientation. The choice

\frac{(x-3)^2}{12}-\frac{(y+1)^2}{3}=1

uses incorrect signs in the binomials: completing the square on

x^2+6x

must produce

x+3

, not

x-3

, and on

-4y^2+8y

must produce

y-1

, not

y+1

. A strong study habit: always complete the square before identifying the conic, and double-check the signs inside the binomials by expanding them back. That one check catches most errors.

Question 11

What is the radius of the circle represented by $2x^2+2y^2-12x+8y+20=0$ ?

$3$
$9$
$\sqrt{10}$
$\sqrt{3}$ (correct answer)

Explanation: Whenever a circle equation is given in expanded general form, your first move should be to complete the square to rewrite it as

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

. Here, divide every term by 2 first:

x^2+y^2-6x+4y+10=0

. Group the

x

and

y

terms and complete the square:

(x^2-6x+9)+(y^2+4y+4)=-10+9+4

. This gives

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

. In standard form, the right side is

r^2

, so

r^2=3

and the radius is

r=\sqrt{3}

. The trap choices come from confusing

r^2

with

r

. The choice

3

is the value of

r^2

, not the radius itself. The choice

9

may tempt you if you square

3

or mistake the constant for the radius. The choice

\sqrt{10}

appears if you forget to move the constant correctly and treat

10

r^2

after completing the square; here the completed-square right side is

3

, not

10

. Study tip: always get the equation into

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

, then take the square root of the right side. On this exam, the most common circle trap is reporting

r^2

instead of

r

, so check that final step.

Question 12

A student divides $2x^2+2y^2-8x+4y+8=0$ by 2 and claims the graph is a circle of radius 2 centered at $(2,-1)$ . What is the actual graph?

A circle centered at $(2,-1)$ with radius 1 (correct answer)
A circle centered at $(2,-1)$ with radius 2
A circle centered at $(-2,1)$ with radius 1
No real points in the plane (the empty set)

Explanation: When you see an equation like

2x^2+2y^2-8x+4y+8=0

, your first instinct should be to check whether it can be written in the standard circle form

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

. Because the coefficients of

x^2

and

y^2

are equal, this is indeed a circle—but you must complete the square to find its true center and radius. Divide the whole equation by 2 first:

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

. Now group and complete the square:

(x^2-4x)+(y^2+2y)+4=0

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

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

.
So the actual graph is a circle centered at

(2,-1)

with radius

\sqrt{1}=1

. The student's claim of radius 2 likely came from mistakenly treating the constant 4 (or the coefficient of the linear term) as the squared radius. A radius of 2 would require

r^2=4

on the right, but our completed square gives 1. The center at

(-2,1)

would arise if you incorrectly used

(x+2)^2

and

(y-1)^2

, but the signs here are negative for

x

and positive for

y

, so the center is indeed

(2,-1)

. Finally, "no real points" would be true only if the right side were negative (a negative radius squared), but here it is positive 1, so the circle exists. Your takeaway: never guess the radius from the constants before completing the square. Always rewrite in standard form, and double-check the signs inside the parentheses to avoid flipping the center.

Question 13

For what value of $c$ does $x^2+y^2-6x+10y+c=0$ represent a single point?

16
34 (correct answer)
64
136

Explanation: Complete the square:

(x-3)^2+(y+5)^2=34-c

. The graph is one point exactly when the right side is 0, so

34-c=0

and

c=34

. If

c<34

the graph is a circle; if

c>34

there are no real points. Choice 16 uses the difference of the linear coefficients, 64 uses half of their sum squared, and 136 uses the linear coefficients without halving them.

Question 14

For a positive value of $a$ , the equation $x^2+y^2-ax+6y+23=0$ has radius 2. What is $a$ ?

$2\sqrt{2}$
$4\sqrt{2}$
$6\sqrt{2}$ (correct answer)
$8\sqrt{2}$

Explanation: Complete the square:

\left(x-\frac{a}{2}\right)^2+(y+3)^2=\frac{a^2}{4}+9-23=\frac{a^2}{4}-14

. Since the radius is 2, the right side is 4:

\frac{a^2}{4}-14=4

, so

a^2=72

and

a=6\sqrt{2}

. The other choices come from using the wrong constant term or solving

\frac{a^2}{4}=2

\frac{a^2}{4}=32

Question 15

For which values of $F$ does $x^2+y^2-6x+4y+F=0$ represent a real, nondegenerate circle?

$F>13$
$F\le 13$
$F<13$ (correct answer)
$F=13$

Explanation: Complete the square:

(x-3)^2+(y+2)^2=13-F

. A real, nondegenerate circle requires the squared-radius quantity to be positive, so

13-F>0

, or

F<13

. If

F=13

, the graph collapses to a point; if

F>13

, it has no real points. The condition

F\le 13

incorrectly includes the point-circle case.

Question 16

How many real points lie on the graph of $x^2+y^2+2x-4y+6=0$ ?

0 (correct answer)
1
2
Infinitely many

Explanation: Completing the square gives

(x+1)^2-1+(y-2)^2-4+6=0

, so

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

. Since the sum of two squares cannot be negative, no real ordered pair satisfies the equation. A zero right side would give a single point, and a positive right side would give a circle with infinitely many real points; a single nondegenerate conic graph cannot consist of exactly two points.

Question 17

A student classifies $x^2+4y^2-6x+8y+13=0$ as an ellipse because completing the square gives $(x-3)^2+4(y+1)^2=0$ . Which statement best identifies the error?

The equation is actually a circle, not an ellipse, because the coefficients of the squared terms are 1 and 4.
The right side must be a positive constant for a nondegenerate ellipse; a right side of 0 makes the graph a single point. (correct answer)
The signs of the squared terms are opposite, so the graph is a hyperbola.
The linear terms cannot be eliminated, so the graph is a parabola.

Explanation: The completed-square form is correct, but a nondegenerate ellipse requires a positive constant on the right. With 0 on the right, the only solution is

(x,y)=(3,-1)

, so the graph is a degenerate point, not an ellipse. The signs are the same, so it is not a hyperbola; unequal coefficients do not make it a circle; and the quadratic terms are still present, so it is not a parabola.

Question 18

Which statement correctly classifies the graph of $4x^2+9y^2-16x+18y-11=0$ ?

It is an ellipse with center $(2,-1)$ and horizontal major axis of length 6. (correct answer)
It is an ellipse with center $(-2,1)$ and horizontal major axis of length 6.
It is a circle with center $(2,-1)$ and radius 3.
It is a hyperbola with center $(2,-1)$ and transverse axis of length 6.

Explanation: Group terms:

4(x^2-4x)+9(y^2+2y)-11=0

. Completing squares gives

4[(x-2)^2-4]+9[(y+1)^2-1]-11=0

, so

4(x-2)^2+9(y+1)^2=36

, or

\frac{(x-2)^2}{9}+\frac{(y+1)^2}{4}=1

. The coefficients have the same sign but are unequal, so the conic is an ellipse, not a circle or hyperbola. The center is

(2,-1)

and the larger denominator is under

x

, so the horizontal major axis has length

2(3)=6

. The other choices have incorrect signs, equal coefficients, or opposite signs.

Question 19

After completing the square, which standard form is equivalent to $x^2-4x-8y+20=0$ ?

$(y-2)^2=8(x-2)$
$(x-2)^2=-8(y-2)$
$(x-2)^2=8(y+2)$
$(x-2)^2=8(y-2)$ (correct answer)

Explanation: Complete the square in

x

(x-2)^2-4-8y+20=0

, so

(x-2)^2=8y-16=8(y-2)

. This is a vertical parabola with vertex

(2,2)

. A form with

(y-2)^2=8(x-2)

would be a horizontal parabola; a negative right side would open downward; and

(x-2)^2=8(y+2)

has the wrong vertical shift.

Question 20

The graph of $x^2+y^2-8x+2y+17=0$ is best described as which of the following?

A circle with center $(4,-1)$ and radius 4.
A circle with center $(-4,1)$ and radius 1.
A single point $(4,-1)$ . (correct answer)
No real points.

Explanation: Complete the square:

(x-4)^2-16+(y+1)^2-1+17=0

, so

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

. A sum of squares equal to 0 is satisfied only at

x=4,\ y=-1

. Thus the degenerate conic is a single point. Choice A incorrectly treats the center coordinate 4 as the radius; choice B has sign errors; choice D would require the right side to be negative.