Unit 3 Test (Lessons 23–30): Introductions to Conics Solutions

#	Answer	Lesson Origin
1A)	Equation G and equation K are hyperbolas because $A \neq C$ are non-zero, and have opposite signs. Equation H is a circle because $A = C$ and has non-zero values. Equation J is a parabola because $C = 0 .$	24, 30
1B)	Equation J is the parabola. A shift left occurs when $(x - h) \to (x - (- 15))$ or $(x + 15) .$ $\frac{x^{2}}{10} - \frac{100}{10} = \frac{10 y}{10} y = \frac{1}{10} {(x)}^{2} - 10 transfomation y = \frac{1}{10} {(x + 15)}^{2} - 10$	27
1C)	$H : x^{2} + (y^{2} + 2 y + {(\frac{2}{2})}^{2}) = 24 + {(\frac{2}{2})}^{2} H : x^{2} + {(y + 1)}^{2} = 25 center (0, - 1), r = 5$ Orange is part C solution Blue is optional sketching to complete part D.	24, 28
1D)	$\begin{matrix} 2 b = 10 & 2 a = 6 \\ b = 5 & a = 3 \\ b^{2} = 25 & a^{2} = 9 \end{matrix} center (0, - 1) \frac{x^{2}}{9} + \frac{{(y + 1)}^{2}}{25} = 1$	29
2)	A	28
3)	C	25
4)	C	23
5)	D	29
6)	B	28, 29
7)	C	24
8)	C	26
9)	A	29
10)	D	30
11)	B	24
12)	B	27
13)	A	23
14)	D	26
15)	C	26, 28
16)	B	24, 25
17)	C	24, 25
18)	D	27
19)	A	30
20)	B	26
21)	D	30
22)	A	23
23)	C	25
24)	$\pm \frac{\sqrt{6}}{2} 3 \pm 2 i$	23, 24
25)	A and C are non-zero A and C have the same signs $A \neq C$	29, 30

Answer all parts of the open response problem.

Use the equations of conic sections to answer parts A–D.

$G : 64 {(y - 4)}^{2} - 25 x^{2} = 1600 J : x^{2} = 10 y + 100$

$H : x^{2} + y^{2} + 2 y = 24 K : 4 x^{2} + 64 x - 25 {(y - 8)}^{2} = - 156$

Name the type of conic each equation represents. Explain your reasoning.

Sample: Equation G and equation K are hyperbolas because $A \neq C,$ are non-zero, and have opposite signs. Equation H is a circle because $A = C$ and has non-zero values. Equation J is a parabola because $C = 0 .$

Write the equation of a parabola in standard form, then transform it left 15 spaces.

Equation J is the parabola. A shift left occurs when $(x - h) \to (x - (- 15))$ or $(x + 15) .$

$\frac{x^{2}}{10} - \frac{100}{10} = \frac{10 y}{10} y = \frac{1}{10} {(x)}^{2} - 10 transfomation y = \frac{1}{10} {(x + 15)}^{2} - 10$

Graph the equation of the circle.

$H : x^{2} + (y^{2} + 2 y + {(\frac{2}{2})}^{2}) = 24 + {(\frac{2}{2})}^{2} H : x^{2} + {(y + 1)}^{2} = 25 center (0, - 1), r = 5$

Orange is part C solution
Blue is optional sketching to complete part D.

Write the equation of an ellipse with the same center and vertical major axis as the circle and with a minor axis equal to 6 units.

$\begin{matrix} 2 b = 10 & 2 a = 6 \\ b = 5 & a = 3 \\ b^{2} = 25 & a^{2} = 9 \end{matrix} center (0, - 1) \frac{x^{2}}{9} + \frac{{(y + 1)}^{2}}{25} = 1$

Multiple Choice

Which equation represents the given conic section?

$\frac{{(x - 2.5)}^{2}}{25} + \frac{{(y + 4.5)}^{2}}{25} = 1$
$\frac{{(x + 2.5)}^{2}}{25} + \frac{{(y - 4.5)}^{2}}{25} = 1$
$\frac{{(x - 2.5)}^{2}}{25} - \frac{{(y + 4.5)}^{2}}{25} = 1$
$\frac{{(x - 4.5)}^{2}}{25} + \frac{{(y + 2.5)}^{2}}{25} = 1$

center $(2.5, - 4.5),$ radius = 5, $r^{2} = 25$

Note

The signs of h and k are opposite the correct value.
This option is a hyperbola because the terms are subtracted.
The value of h and k are switched in the equation.

An object launched into the air is modeled by the equation: $y = - 16 t^{2} + 144 t + 160$ . How many seconds, t, after launch does the object reach the ground?

–1
2
10
16

This equation can be solved by factoring, using the Quadratic Formula, or completing the square.

$\begin{array}{rcl} - 16 t^{2} - 144 t - 160 & = & 0 \\ - 16 (t^{2} - 9 t - 10) & = & 0 \\ - 16 (t - 10) (t + 1) & = & 0 \\ t & = & 10, - 1 \end{array}$

Note

Time cannot be negative.
This option is the result when 2 and 5 are used as factors of 10 instead of 1 and 10.
This option is the result if the GCF is not factored out.

Solve $x^{2} + 12 = 0$ under the set of complex numbers.

$2 i \sqrt{3}$
$2 \sqrt{3}$
$\pm 2 i \sqrt{3}$
no solution

$x^{2} = - 12 \sqrt{x^{2}} = \pm \sqrt{- 12} x = \pm 2 i \sqrt{3}$

Note

This option is only one of two possible solutions.
With this option, the square root of a negative number will be imaginary.
Because complex numbers are being used, a solution is possible.

Select the equation that translates the center of an ellipse at the origin into the third quadrant.

$\frac{{(x + 13)}^{2}}{5} - \frac{{(y + 8)}^{2}}{7} = 1$
$\frac{{(x - 13)}^{2}}{5} + \frac{{(y - 8)}^{2}}{7} = 1$
$\frac{{(x - 13)}^{2}}{5} + \frac{{(y + 8)}^{2}}{7} = 1$
$\frac{{(x + 13)}^{2}}{5} + \frac{{(y + 8)}^{2}}{7} = 1$

center $(0, 0)$ translated to the third quadrant $(- h, - k)$

$\frac{{(x - (- h))}^{2}}{a^{2}} + \frac{{(y - (- k))}^{2}}{b^{2}} = 1 \frac{{(x + h)}^{2}}{a^{2}} + \frac{{(y + k)}^{2}}{b^{2}} = 1$

Note

This option is a hyperbola because the terms are subtracted.
This option is an ellipse with a center in the first quadrant.
This option is an ellipse with a center in the fourth quadrant.

Select the statement that is true.

An ellipse is a special circle where $a = b .$
A circle is a special ellipse where $a = b .$
A circle is a special ellipse where $a \neq b .$
There is no relationship between a circle and an ellipse.

Note

The values of a and b are equal when the equation of a circle is written in the standard form of an ellipse.

What value of b will form a perfect square trinomial $x^{2} + b x = - 14$ ?

7
$\sqrt{14}$
$2 \sqrt{14}$
49

$x^{2} + b x + 14 = 0 b = 2 \sqrt{a c}, a = 1, c = 14 b = 2 \sqrt{(1) (14)} b = 2 \sqrt{14}$

Note

This option is half of c.
This option does not double the middle term of the trinomial.
This option is the correct value if you were asked for c.

Determine the distance between point A and the midpoint of segment AB.

5.57
4.72
2.36
1.18

$A (3, - 4), B (- 1, - 1.5) midpoint M M (\frac{3 + - 1}{2}, \frac{- 4 + - 1.5}{2}) = (1, - 2.75) A to M d = \sqrt{{(3 - 1)}^{2} + {(- 4 - (- 2.75))}^{2}} d = \sqrt{{(2)}^{2} + {(- 1.25)}^{2}} d = 2.358 \approx 2.36$

Note

This option is the approximate value before taking the square root.
This option is the distance between A and B.
This option is half the distance between the midpoint and point A.

Select the equation that best represents the graph.

$\frac{{(y - 2)}^{2}}{4} + \frac{{(x + 3)}^{2}}{25} = 1$
$\frac{{(y - 2)}^{2}}{4} - \frac{{(x + 3)}^{2}}{25} = 1$
$\frac{{(y - 2)}^{2}}{4} + \frac{{(x - 3)}^{2}}{25} = 1$
$\frac{{(y - 2)}^{2}}{25} + \frac{{(x + 3)}^{2}}{4} = 1$

Center $(- 3, 2)$

Major axis horizontal, $2 a = 10, a = 5$

$2 b = 4, b = 2$

Note

This option equation represents a hyperbola.
This option represents an ellipse with a center in quadrant one.
This option is a vertical ellipse.

Solve $x^{2} + 38 = 2 x$ by completing the square.

$\pm i \sqrt{37}$
$1 + i \sqrt{37}$
$1 \pm \sqrt{37}$
$1 \pm i \sqrt{37}$

$x^{2} - 2 x = - 38 x^{2} - 2 x + {(- \frac{2}{2})}^{2} = - 38 + {(- \frac{2}{2})}^{2} {(x - 1)}^{2} = - 38 + 1 {(x - 1)}^{2} = - 37 \sqrt{{(x - 1)}^{2}} = \pm \sqrt{- 37} x - 1 = \pm i \sqrt{37} x = 1 \pm i \sqrt{37}$

Note

This option does not add 1 to both sides of the equation.
This option represents only one of the two solutions.
This option is missing i in the solution.

Which equation represents a hyperbola?

$7 x^{2} + y^{2} + 6 y = 5 x + y^{2}$
$8 y^{2} + 6 y + 12 = 7 x^{2} + y^{2} + 5 x$
$7 x^{2} + 8 y^{2} + 6 y = y^{2} + 5 x + 12$
$7 y^{2} + 6 y + 12 = 7 x^{2} + 8 y^{2} + 5 x$

Note

This option is a parabola because A is non-zero and $C = 0 .$
This option is a circle because $A = C .$
This option is an ellipse because A and C are non-zero values with the same sign.

Select the graph that best represents the equation: $x = 3 {(y - 2)}^{2} - 1$ .

$a = 3 (h, k) : (- 1, 2)$

Note

A, D) These options have a vertex at $(- 2, 1) .$

C, D) These options have $a = \frac{1}{3}$ rather than 3.

Which equation represents a polynomial equation with integer coefficients given the roots $x = 1, \sqrt{13}$ ?

$x^{3} - x^{2} - 13 x + 13 = 0$
$x^{2} - x - x \sqrt{13} + \sqrt{13}$
$x^{3} + x^{2} - 13 x - 13 = 0$
$x^{3} - 12 x + 13 = 0$

$\begin{array}{rcl} \begin{matrix} x = - \sqrt{13}, & x = \sqrt{13}, & x = 1 \end{matrix} \\ \begin{matrix} (x - (- \sqrt{13})) = 0, & (x - \sqrt{13}) = 0, & x - 1 = 0 \end{matrix} (x - (- \sqrt{13})) (x - \sqrt{13}) (x - 1) = 0 (x + \sqrt{13}) (x - \sqrt{13}) (x - 1) = 0 (x^{2} x^{2} - x \sqrt{13} + x \sqrt{13} - \sqrt{169}) (x - 1) = 0 (x^{2} - 13) (x - 1) = 0 \end{array}$

Note

This option does not have integer coefficients and is not an equation.
This option is the solution when $(x + 1)$ is used as a factor.
This option incorrectly combines the squared and linear terms.

Determine the distance between the vertices of the two parabolas $x = - {(y - 2.25)}^{2} + 5$ and $y = - {(x + 4)}^{2} + 3.4$ to the nearest hundredth.

1.60
5.74
8.93
9.07

$(- 4, 3.4), (5, 2.25) d = \sqrt{{(- 4 - 5)}^{2} + {(3.4 - 2.25)}^{2}} d = 9.0731 \approx 9.07$

Note

This option is the difference between the constant terms.

B–C) These values occur when the addition and subtraction symbols are incorrectly placed.

Write the equation of a circle with endpoints on the diameter at $(18, 11)$ and $(- 19, - 13) .$

${(x - 0.5)}^{2} + {(y - 1)}^{2} = 486.25$
${(x + 1)}^{2} + {(y + 0.5)}^{2} = 486.25$
${(x + 0.5)}^{2} + {(y + 1)}^{2} = 486.25$
${(x + 0.5)}^{2} + {(y + 1)}^{2} = 1945$

$(18, 11), (- 19, - 13) midpoint : center (\frac{18 + - 19}{2}, \frac{11 + - 13}{2}) center (- \frac{1}{2}, - 1) r^{2} = {(\sqrt{{(- 0.5 - 18)}^{2} + {(- 1 - 11)}^{2}})}^{2} r^{2} = 486.25$

Note

This option has the opposite values for h and k.
The values of h and k are switched in the equation.
This option uses the length of the diameter rather than the radius.

Order the steps to correctly show the derivation of the Quadratic Formula from the equation $a x^{2} + b x + c = 0 .$

I, II, III, IV
II, I, IV, III
III, IV, I, II
IV, I, II, III

$I \sqrt{{(x + \frac{b}{2 a})}^{2}} = \pm \sqrt{\frac{- 4 a c + b^{2}}{4 a^{2}}} II x^{2} + \frac{b}{a} x + \frac{b^{2}}{4 a^{2}} = - \frac{c}{a} + \frac{b^{2}}{4 a^{2}} III x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} IV x + \frac{b}{2 a} = \frac{\pm \sqrt{- 4 a c + b^{2}}}{2 a}$

What type of roots will result from the equation $x^{2} - 6 x + 9 = 8 ?$

one real, rational root
two real, rational roots
two real, irrational roots
two imaginary roots

$x^{2} - 6 x = - 1 x^{2} - 6 x + {(\frac{- 6}{2})}^{2} = - 1 + {(\frac{- 6}{2})}^{2} {(x - 3)}^{2} = - 1 + {(- 3)}^{2} \sqrt{{(x - 3)}^{2}} = \pm \sqrt{8} x - 3 = \pm 2 \sqrt{2} x = 3 \pm 2 \sqrt{2}$

Note

A, B) The answer cannot be rational when it includes the square root of a prime number.

The square root of a negative does not occur, so the answer is not imaginary.

Determine the graph of the parabola that opens left and is wider than the graph of $x = y^{2} .$

$x = 3 y^{2} + 2 x - 5$
$x = \frac{1}{3} y^{2} + 2 x - 5$
$x = - 3 y^{2} + 2 x - 5$
$x = - \frac{1}{3} y^{2} + 2 x - 5$

$0 < | a | < 1$

The parabola is wider than the parent graph.

When $a = a$ is a negative value, the graph opens left.

Note

A, C) When $| a | > 1,$ the parabola is narrower.
A, B) When $a = a$ is a positive value, the parabola opens right.

Which statement describes the graph of the equation $3 x^{2} - 2 y^{2} + 12 x + 8 y + 6 = 0$ ?

a hyperbola with center $(- 2, 2)$ and vertices $(- 2, 1)$ and $(- 2, 3)$
a hyperbola with center $(2, - 2)$ and vertices $(2, - 1)$ and $(2, - 3)$
an ellipse with center $(2, - 2)$ and a major axis length of 2
an ellipse with center $(- 2, 2)$ and a major axis length of 4

$3 x^{2} + 12 x - 2 y^{2} + 8 y = - 6 3 (x^{2} + 4 x + {(\frac{4}{2})}^{2}) - 2 (y^{2} - 4 y + {(\frac{- 4}{2})}^{2}) = - 6 + 3 {(\frac{4}{2})}^{2} - 2 {(\frac{- 4}{2})}^{2} 3 {(x + 2)}^{2} - 2 {(y - 2)}^{2} = - 6 + 3 (4) - 2 (4) \frac{3 {(x + 2)}^{2}}{- 2} - \frac{2 {(y - 2)}^{2}}{- 2} = \frac{- 2}{- 2} \frac{3 {(x + 2)}^{2}}{- 2} + {(y - 2)}^{2} = 1$

Note

B, C) The signs of the value of the center are reversed.
C, D) The given equation represents a hyperbola.

Determine the length of segment AB with the midpoint $M (2.5, 3.5)$ and point $A (8, 1) .$

11.42
12.08
6.04
0

$\begin{matrix} \frac{8 + x}{2} = 2.5 & \frac{1 + y}{2} = 3.5 \\ 8 + x = 5 & 1 + y = 7 \\ x = - 3 & y = 6 \end{matrix} B (- 3, 6) A (8, 1) d = \sqrt{{(- 3 - 8)}^{2} + {(6 - 1)}^{2}} d = 12.083$

Note

This option is the value when only addition is used in the distance formula.
This option is the distance from the midpoint to an endpoint 6.04.
This option is the value when only subtraction is used in the distance formula.

Determine the set of equations that represents the asymptotes of the hyperbola: $\frac{{(x - 1)}^{2}}{9} - \frac{{(y + 3)}^{2}}{4} = 1$

$y = \frac{3}{2} (x - 1) + 3, y = - \frac{3}{2} (x - 1) + 3$
$y = \frac{2}{3} (x - 1) + 3, y = - \frac{2}{3} (x - 1) + 3$
$y = \frac{3}{2} (x - 1) - 3, y = - \frac{3}{2} (x - 1) - 3$
$y = \frac{2}{3} (x - 1) - 3, y = - \frac{2}{3} (x - 1) - 3$

$center (1, - 3), a = 3, b = 2 y - k = m (x - h) + k + k y = \pm m (x - h) + k y = \pm \frac{b}{a} (x - h) + k$

Note

A, B) These options move the asymptote up three, rather than down three spaces.

A, C) These options have the incorrect solution for the asymptotes.

If $(4 + i)$ is a solution for $x^{2} + b x + c = 0$ where b and c are real numbers, what is the value of c ?

17
16
–1
–8

$x = 4 + i, 4 - i (x - (4 + i)) (x - (4 - i)) = 0 (x - 4 - i) (x - 4 + i) = 0 x^{2} - 4 x + i x - 4 x + 16 - 4 i - i x + 4 i - i^{2} = 0 x^{2} - 8 x + 16 - (- 1) = 0 x^{2} - 8 x + 17 = 0$

Note

This option is the value if $i^{2}$ is dropped from the problem.
This option is the value of $i^{2}$ when simplified.
This option is the value of b.

Determine the quadratic equation with exactly one real solution.

$x^{2} - 3 x + 9 = 0$
$2 x^{2} + 2 x = 0$
$4 x^{2} - 12 x + 9 = 0$
$x^{2} + 3 x + 7 = 0$

$\begin{array}{lll} A & B \\ a = 1, b = - 3, c = 9 & a = 2, b = 2, c = 0 \\ x = \frac{- (- 3) \pm \sqrt{{(- 3)}^{2} - 4 (1) (9)}}{2 (1)} & x = \frac{- (2) \pm \sqrt{{(2)}^{2} - 4 (2) (0)}}{2 (2)} \\ x = \frac{3 \pm \sqrt{9 - (36)}}{2} & x = \frac{- 2 \pm \sqrt{4 - 0}}{4} \\ x = \frac{3 \pm \sqrt{- 27}}{2} & x = \frac{- 2 \pm \sqrt{4}}{4} \\ x = \frac{3 \pm 3 i \sqrt{3}}{2} & x = - 1, 0 \end{array}$

$\begin{array}{lcl} C & D \\ {(2 x - 3)}^{2} = 0 & a = 1, b = 3, c = 7 \\ 2 x - 3 = 0 & x = \frac{- (3) \pm \sqrt{{(3)}^{2} - 4 (1) (7)}}{2 (1)} \\ 2 x = 3 & x = \frac{- 3 \pm \sqrt{9 - (28)}}{2} \\ x = \frac{3}{2} & x = \frac{- 3 \pm i \sqrt{19}}{2} \end{array}$

Note

A, D) These options have two imaginary complex solutions.

This option has two real solutions.

Select all possible solutions under the set of complex numbers for the polynomial equation $(2 x^{2} - 3) (x^{2} - 6 x + 13) = 0 .$ Remember to rationalize solutions.

$\pm \sqrt{3}$
$\pm \frac{\sqrt{6}}{2}$
1, 5
$\pm 2 i$
$3 \pm 2 i$

Note

The incorrect options do not result in the equation being equal to zero when solved.

$\begin{array}{rcl} \begin{matrix} 2 x^{2} = 3 & x^{2} - 6 x = - 13 \\ \sqrt{x^{2}} = \pm \sqrt{\frac{3}{2}} & x^{2} - 6 x + {(\frac{- 6}{2})}^{2} = - 13 + {(\frac{- 6}{2})}^{2} \\ x = \pm \frac{\sqrt{3}}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} & {(x - 3)}^{2} = - 13 + 9 \\ x = \pm \frac{\sqrt{6}}{2} & \sqrt{{(x - 3)}^{2}} = \pm \sqrt{- 4} \\ x - 3 = \pm 2 i \\ x = 3 \pm 2 i \end{matrix} \end{array}$

Select the rules to determine an ellipse in the general form $A x^{2} + B x y + C y^{2} + D x + E y + F = 0 .$

$A = C$
A and C are non-zero
A and C have opposite signs
A and C have the same signs
$A \neq C$

Note

“A = C ”: For an ellipse, A cannot equal C.

“A and C have opposite signs”: For an ellipse, A and C cannot have opposite signs.

If $A \neq C$ and are non-zero and have the same sign, the conic is an ellipse.