Question 1

Graph the equation 2 x ² + 5 y ² = 0. A) B) C) D)

Accepted Answer

The equation can be simplified by factoring out a common term of 2^{:
2^{(1 + 5y}) = 0
This has two solutions:
- 2^{= 0 (which is equivalent to ^{= 0, since we can divide both sides by 2)
- 1 + 5y} = 0, which can be rearranged as:
y = -1 / (5})
This is the equation of a horizontal line with slope 0 and y-intercept -1 / (5}). The graph of the full equation consists of the two lines y = 0 (which corresponds to ^{= 0) and y = -1 / (5}), as shown in choice C.

Question 2

Graph the hyperbola 16 x 2 $-$ 9 y 2 = 16.
A) 
B) 
C) 
D)

Accepted Answer

The answer of Graph the hyperbola 16 x 2 $-$...

Question 3

Graph the hyperbola x 2 $-$ y 2 = 2.
A) 
B) 
C) 
D)

Accepted Answer

The equation x^2 - y^2 = 2 is already in standard form (with the x-term being positive and the y-term being negative), so we know that this is a hyperbola centered at (0,0). To graph it, we can use the fact that the vertices are located at (sqrt(2),0) and (-sqrt(2),0), and the asymptotes have equations y = x and y = -x. Plugging in some points around the vertices and along the asymptotes, we can sketch the hyperbola. Choice B is the closest to our sketch.

Question 4

Graph the hyperbola. $x ^ { 2 } - 6 x - y ^ { 2 } + 4 y - 11 = 0$
A) 
B) 
C) 
D) 
E)

Accepted Answer

The answer of Graph the hyperbola. \(x ^ { 2...

Question 5

Graph the hyperbola. $9 x ^ { 2 } - 36 x - 4 y ^ { 2 } = 0$
A) 
B) 
C) 
D) 
E)

Accepted Answer

The answer of Graph the hyperbola. \(9 x ^ {...

Question 6

The graphs of equations of the form xy = k , where k is a nonzero constant, are also hyperbolas, sometimes referred to as rectangular hyperbolas. Graph the hyperbola xy = 15.
A) 
B) 
C) 
D)

Accepted Answer

The answer of The graphs of equations of the form...

Question 7

Graph the hyperbola 9 y 2 $-$ 18 y $-$ 16 x 2 = 0.
A) 
B) 
C) 
D)

Accepted Answer

The answer of Graph the hyperbola 9 y 2 $-$...

Question 8

The graphs of equations of the form xy = k , where k is a nonzero constant, are also hyperbolas, sometimes referred to as rectangular hyperbolas. Graph the hyperbola xy = 1.
A) 
B) 
C) 
D)

Accepted Answer

The answer of The graphs of equations of the form...

Question 9

Graph the hyperbola $-$ 9 x 2 + 72 x + 4 y 2 $-$ 8 y $-$ 176 = 0.
A) 
B) 
C) 
D)

Accepted Answer

The answer of Graph the hyperbola $-$ 9 x 2...

Question 10

Graph the hyperbola y 2 $-$ 16 x 2 = 9.
A) 
B) 
C) 
D)

Accepted Answer

First, we need to rewrite the equation in standard form, which is:
 
    (y^2/9) - (x^2/16) = 1
 
We can see that the transverse axis is the y-axis since y^2 is positive and x^2 is negative. The center is (0,0) and the distance from the center to the vertices is 3 units. Hence, the vertices are at (0, ±3).
 
The distance from the center to the foci is given by c = sqrt(a^2 + b^2), where a=3 and b=4. Solving for c gives us c=5. Therefore, the foci are at (0, ±5).
 
The asymptotes can be found using the formula y = ±(b/a)x. Substituting a=3 and b=4, we get the equations y = ±(4/3)x.
 
The graph of the hyperbola should look like this:
 
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Question 11

Graph the hyperbola $-$ 9 x 2 + 36 x + 4 y 2 + 24 y $-$ 16 = 0.
A) 
B) 
C) 
D)

Accepted Answer

The answer of Graph the hyperbola $-$ 9 x 2...

Question 12

Graph the hyperbola $-$ 4 x 2 + y 2 = $-$ 4.
A) 
B) 
C) 
D)

Accepted Answer

The answer of Graph the hyperbola $-$ 4 x 2...

Question 13

Find the intercepts for $9 x ^ { 2 } - 4 y ^ { 2 } = 36$ .

Accepted Answer

The answer of Find the intercepts for \(9 x ^...

Question 14

Graph the hyperbola x 2 + 8 x $-$ 9 y 2 + 72 y $-$ 144 = 0.
A) 
B) 
C) 
D)

Accepted Answer

First, we need to rearrange the given equation into standard form:

(x^2 + 8x) - 9(y^2 - 8y) = 144

(x^2 + 8x + 16) - 9(y^2 - 8y + 16/9) = 144 + 16 - 16/9

(x + 4)^2/16 - 9(y - 4/3)^2/16 = 157/9

Now, we can see that this hyperbola has its center at (-4, 4/3), and its transverse axis is along the x-axis with a length of 4(sqrt(157))/3. Its conjugate axis is along the y-axis with a length of 4(sqrt(157))/9.

To graph the hyperbola, we can start by plotting the center point (-4, 4/3). Then, we can draw the transverse axis along the x-axis with a length of 4(sqrt(157))/3, which means we need to plot the points (-4 - 2(sqrt(157))/3, 4/3) and (-4 + 2(sqrt(157))/3, 4/3).

Next, we can draw the asymptotes, which will pass through the center point (-4, 4/3) and have a slope of +/- b/a, where b is the length of the conjugate axis and a is the length of the transverse axis. So, the slopes of the asymptotes are +/- (4/3)/(4(sqrt(157))/3) = +/- 1/sqrt(157).

Using these slopes and the center point (-4, 4/3), we can plot the asymptotes and extend them until they intersect the coordinate axes.

Finally, we can sketch the hyperbola by drawing its two branches, which will intersect the transverse axis at the points we plotted earlier and will approach the asymptotes as they extend outward.

The resulting graph should match choice C.

Question 15

The graphs of equations of the form xy = k , where k is a nonzero constant, are also hyperbolas, sometimes referred to as rectangular hyperbolas. Graph the hyperbola xy = $-$ 2.
A) 
B) 
C) 
D)

Accepted Answer

We can rewrite xy = $-$ 2 as y = $-$ $\frac{2}{x}$. This equation represents a hyperbola with vertical and horizontal asymptotes. As x approaches 0, y becomes infinitely large in the negative and positive direction. As y approaches 0, x becomes infinitely large in the negative and positive direction. Choice C is the only graph that shows both sets of asymptotes, indicating that it is the correct graph for the hyperbola xy = $-$ 2.

Question 16

Graph the hyperbola 4 y 2 $-$ x 2 = 9.
A) 
B) 
C) 
D)

Accepted Answer

The answer of Graph the hyperbola 4 y 2 $-$...

Question 17

Graph the hyperbola 4 x 2 $-$ y 2 = 20.
A) 
B) 
C) 
D)

Accepted Answer

First, we need to rearrange the given equation into the standard form of a hyperbola, which is:

((x-h)^2/a^2) - ((y-k)^2/b^2) = 1

where (h,k) is the center of the hyperbola.

So, starting with the given equation:

4x^2 - y^2 = 20

Dividing both sides by 20, we get:

(x^2/5) - (y^2/20) = 1

Comparing this to the standard form above, we see that:

h = 0, k = 0, a^2 = 5, b^2 = 20

Therefore, the center of the hyperbola is at (0,0) and the distance from the center to each vertex is a = sqrt(5).

We also know that the foci, which are the points (h ± c, k), are located a distance of c = sqrt(a^2 + b^2) from the center.

So, plugging in the values for a and b, we get:

c = sqrt(5 + 20) = sqrt(25) = 5

Therefore, the foci are located at (-5,0) and (5,0).

To graph the hyperbola, we start by drawing the x and y axes and marking the center at (0,0).

Then, we draw the transverse axis, which is the line passing through the center and the two vertices, which are (sqrt(5),0) and (-sqrt(5),0). This axis has a length of 2a = 2sqrt(5) and is shown in blue in the graph.

Next, we draw the two branches of the hyperbola, which are mirror images of each other across the transverse axis. Since we know that the foci are located a distance of 5 from the center, we can draw a rectangle with sides of length 2a and 2c, with the center of the rectangle at (0,0) and the foci located at (-5,0) and (5,0). Then, we can draw the hyperbola passing through the corners of the rectangle. The branches of the hyperbola are shown in red in the graph.

Therefore, the correct graph is option C.

Question 18

Graph the hyperbola 2 y 2 $-$ 25 x 2 = 50.
A) 
B) 
C) 
D)

Accepted Answer

To graph the hyperbola, we first need to put it into standard form:

2y^2/50 - 25x^2/50 = 1

Simplifying:

y^2/25 - x^2/2 = 1

This tells us that the center of the hyperbola is at (0,0), it opens horizontally, and the distance from the center to the vertex is sqrt(25) = 5.

To graph it, we can plot the center at (0,0) and then plot the vertices at (5,0) and (-5,0). We can also plot the asymptotes, which are lines that pass through the center and have slopes of plus or minus the ratio of the coefficients of x and y in the standard form equation.

The slope of the asymptotes is plus or minus (sqrt(25/2)/5), which simplifies to plus or minus sqrt(2)/2. So we can draw two lines passing through the center with slopes of plus or minus sqrt(2)/2.

Putting all of this together gives us choice B.

Question 19

Graph the equation 3 x 2 $-$ 6 y 2 = 0.
A) 
B) 
C) 
D)

Accepted Answer

The answer of Graph the equation 3 x 2 $-$...

Question 20

Graph the hyperbola y 2 $-$ 9 x 2 = 16.
A) 
B) 
C) 
D)

Accepted Answer

The answer of Graph the hyperbola y 2 $-$ 9...

Quiz 12: Coordinate Geometry: Lines, Parabolas, Circles, Ellipses and Hyperbolas

Graph the equation 2 x 2 + 5 y 2 = 0.

Graph the hyperbola 16 x 2 −-− 9 y 2 = 16.

Graph the hyperbola x 2 −-− y 2 = 2.

Graph the hyperbola. x2−6x−y2+4y−11=0x ^ { 2 } - 6 x - y ^ { 2 } + 4 y - 11 = 0x2−6x−y2+4y−11=0

Graph the hyperbola. 9x2−36x−4y2=09 x ^ { 2 } - 36 x - 4 y ^ { 2 } = 09x2−36x−4y2=0

The graphs of equations of the form xy = k , where k is a nonzero constant, are also hyperbolas, sometimes referred to as rectangular hyperbolas. Graph the hyperbola xy = 15.

Graph the hyperbola 9 y 2 −-− 18 y −-− 16 x 2 = 0.

The graphs of equations of the form xy = k , where k is a nonzero constant, are also hyperbolas, sometimes referred to as rectangular hyperbolas. Graph the hyperbola xy = 1.

Graph the hyperbola −-− 9 x 2 + 72 x + 4 y 2 −-− 8 y −-− 176 = 0.

Graph the hyperbola y 2 −-− 16 x 2 = 9.

Graph the hyperbola −-− 9 x 2 + 36 x + 4 y 2 + 24 y −-− 16 = 0.

Graph the hyperbola −-− 4 x 2 + y 2 = −-− 4.

Find the intercepts for 9x2−4y2=369 x ^ { 2 } - 4 y ^ { 2 } = 369x2−4y2=36 .

Graph the hyperbola x 2 + 8 x −-− 9 y 2 + 72 y −-− 144 = 0.

The graphs of equations of the form xy = k , where k is a nonzero constant, are also hyperbolas, sometimes referred to as rectangular hyperbolas. Graph the hyperbola xy = −-− 2.

Graph the hyperbola 4 y 2 −-− x 2 = 9.

Graph the hyperbola 4 x 2 −-− y 2 = 20.

Graph the hyperbola 2 y 2 −-− 25 x 2 = 50.

Graph the equation 3 x 2 −-− 6 y 2 = 0.

Graph the hyperbola y 2 −-− 9 x 2 = 16.

Graph the equation 2 x ² + 5 y ² = 0.

Graph the hyperbola 16 x 2 $-$ 9 y 2 = 16.

Graph the hyperbola x 2 $-$ y 2 = 2.

Graph the hyperbola. $x ^ { 2 } - 6 x - y ^ { 2 } + 4 y - 11 = 0$

Graph the hyperbola. $9 x ^ { 2 } - 36 x - 4 y ^ { 2 } = 0$

Graph the hyperbola 9 y 2 $-$ 18 y $-$ 16 x 2 = 0.

Graph the hyperbola $-$ 9 x 2 + 72 x + 4 y 2 $-$ 8 y $-$ 176 = 0.

Graph the hyperbola y 2 $-$ 16 x 2 = 9.

Graph the hyperbola $-$ 9 x 2 + 36 x + 4 y 2 + 24 y $-$ 16 = 0.

Graph the hyperbola $-$ 4 x 2 + y 2 = $-$ 4.

Find the intercepts for $9 x ^ { 2 } - 4 y ^ { 2 } = 36$ .

Graph the hyperbola x 2 + 8 x $-$ 9 y 2 + 72 y $-$ 144 = 0.

The graphs of equations of the form xy = k , where k is a nonzero constant, are also hyperbolas, sometimes referred to as rectangular hyperbolas. Graph the hyperbola xy = $-$ 2.

Graph the hyperbola 4 y 2 $-$ x 2 = 9.

Graph the hyperbola 4 x 2 $-$ y 2 = 20.

Graph the hyperbola 2 y 2 $-$ 25 x 2 = 50.

Graph the equation 3 x 2 $-$ 6 y 2 = 0.

Graph the hyperbola y 2 $-$ 9 x 2 = 16.