Graphs - College Algebra

Card 0 of 20

Question

Find the focal points of the conic below:

$27x^{^{2}}+16y^{^{2}}+324x-96y+684 = 0$

Answer

The first thing we want to do is put the conic (an ellipse because the x2 and the y2 terms have the same sign) into a better form i.e.

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

where (h,k) is the center of our ellipse.

We will continue by completing the square for both the x and y binomials.

First we seperate them into two trinomials:

$\left ( 27x^2+324x+0\right )+\left ( 16y^2-96y+0\right )=-684$

then we pull a 27 out of the first one and a 16 out of the second

$27\left ( x^2+12x+0\right )+16\left ( y^2-6y+0\right )=-684$

then we add the correct constant to each trinomial (being sure to add the same amount to the other side of our equation.

$27\left ( x^2+12x+36\right )+16\left ( y^2-6y+9\right )=-684+27(36)+16(9)$

then we factor our trinomials and divide by 16 and 27 to get

$\frac{\left ( x+6\right )^2}{16}+\frac{\left ( y-3\right )^2}{27}=1$

so the center of our ellipse is (-6,3) and we calculate the distance from the focal points to the center by the equation:

$c^2=\left | a^2-b^2\right |$

and we know that our ellipse is stretched in the y direction because b>a so our focal points will be c displaced from our center.

with $c=\sqrt{11}$

our focal points are $\left ( -6,3+\sqrt{11}\right ) , \left (-6,3-\sqrt{11} \right )$

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Question

What is the equation of the elipse centered at the origin and passing through the point (5, 0) with major radius 5 and minor radius 3?

Answer

The equation of an ellipse is

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

where a is the horizontal radius, b is the vertical radius, and (h, k) is the center of the ellipse. In this case we are told that the center is at the origin, or (0,0), so both h and k equal 0. That brings us to:

$\frac{(x-0)^2}{a^2} + \frac{(y-0)^2}{b^2} = 1$

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

We are told about the major and minor radiuses, but the problem does not specify which one is horizontal and which one vertical. However it does tell us that the ellipse passes through the point (5, 0), which is in a horizontal line with the center, (0, 0). Therefore the horizontal radius is 5.

The vertical radius must then be 3. We can now plug these in:

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

$\frac{x^2}{25} + \frac{y^2}{9} = 1$

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Question

An ellipse is centered at (-3, 2) and passes through the points (-3, 6) and (4, 2). Determine the equation of this eclipse.

Answer

The usual form for an ellipse is

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

where (h, k) is the center of the ellipse, a is the horizontal radius, and b is the vertical radius.

Plug in the coordinate pair:

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

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

Now we have to find the horizontal radius and the vertical radius. Let's compare points; we are told the ellipse passes through the point (-3, 6), which is vertically aligned with the center. Therefore the vertical radius is 4.

Similarly, the ellipse passes through the point (4, 2), which is horizontally aligned with the center. This means the horizontal radius must be 7.

Substitute:

$\frac{(x+3)^2}{7^2} + \frac{(y-2)^2}{4^2} = 1$

$\frac{(x+3)^2}{49} + \frac{(y-2)^2}{16} = 1$

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Question

Find the center of this ellipse:

$33x^2+13y^2-396x+26\sqrt{7}y+850=0$

Answer

To find the center of this ellipse we need to put it into a better form. We do this by rearranging our terms and completing the square for both our y and x terms.

$33(x^2-12x)+13(y^2+2\sqrt7y)=-850$

completing the square for both gives us this.

$33(x^2-12x+36)+13(y^2+2\sqrt7y+7)=429$

$33(x-6)^2+13(y+\sqrt7)^2=429$

we could divide by 429 but we have the information we need. The center of our ellipse is $(6,-\sqrt7 )$

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Question

Find the eccentricity of an ellipse with the following equation:

Answer

Start by putting this equation in the standard form of the equation of an ellipse:

$\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$ , where is the center of the ellipse.

Group the terms together and the terms together.

Factor out from the terms and from the terms.

Now, complete the squares. Remember to add the same amount to both sides!

Subtract from both sides.

Divide both sides by .

$\frac{(x^2-16x+64)}{4}+\frac{y^2-2y+1}{25}=1$

Factor both terms to get the standard form of the equation of an ellipse.

$\frac{(x-8)^2}{4}+\frac{(y+1)^2}{25}=1$

Recall that the eccentricity is a measure of the roundness of an ellipse. Use the following formula to find the eccentricity, .

$e=\frac{\text{Distance from center to focus}}{\text{Distance from center to vertex}}$

Next, find the distance from the center to the focus of the ellipse, . Recall that when , the major axis will lie along the -axis and be horizontal and that when , the major axis will lie along the -axis and be vertical.

is calculated using the following formula:

$c=\sqrt{a^2-b^2}$ for , or

$c=\sqrt{b^2-a^2}$ for

For the ellipse in question,

$c=\sqrt{25-4}=\sqrt{21}$

Now that we have found the distance from the center to the foci, we need to find the distance from the center to the vertex.

Because , the major axis for this ellipse is vertical. will be the distance from the center to the vertices.

For this ellipse, .

Now, plug in the distance from the center to the focus and the distance from the center to the vertex to find the eccentricity of this ellipse.

$e=\frac{\sqrt{21}}{5}$

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Question

Write the equation for an ellipse with center , foci $(1 \pm \sqrt{14}, 4)$ and a major axis with length 14.

Answer

The general equation for an ellipse is $\frac{(x-h)^2 }{a^2 } + \frac{(y-k)^2 }{b^2 } = 1$ , although if we consider a to be half the length of the major axis, a and b might switch depending on if the longer major axis is horizontal or vertical. This general equation has as the center, a as the length of half the major axis, and b as the length of half the minor axis.

Because the center of this ellipse is at and the foci are at $(1 \pm \sqrt{14} , 4 )$ , we can see that the foci are $\sqrt{14}$ away from the center, and they are on the horizontal axis. This means that the horizontal axis is the major axis, the one with length 14. Having a length of 14 means that half is 7, so . Since the foci are $\sqrt{14}$ away from the center, we know that $c = \sqrt{14}$ . We can solve for b using the equation :

$7^2 - \sqrt{14}^2 = b^2$

that's really as far as we need to solve.

Putting all this information into the equation gives:

$\frac{(x-1)^2 }{ 49} + \frac{(y - 4)^2 }{35 } = 1$

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Question

What is the equation of the ellipse given the following:

Vertices: (10,0), (-10,0)

Co-vertices: (0,7), (0,-7)

Answer

Standard equation of an ellipse: $\frac{(x-h)^{2}}{a^{2}}+\frac{(y-k)^{2}}{b^{2}}=1$

From the given information, the ellipse is centered around the origin (0,0), so h and k are both 0.

The coordinates of the vertices are on the x-axis, which is the major axis. The vertices are a units away from the center. Here, a = 10.

The coordinates of the co-vertices are on the y-axis, which is the minor axis. The co-vertices are b units away from the center. Here, b = 7.

$\frac{x^{2}}{100}+\frac{y^{2}}{49}=1$

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Question

The graph of the equation

$16x^{2} + 12y^{2} -96x-48y =0$

is an example of which conic section?

Answer

The quadratic coefficients in this general form of a conic equation are 16 and 12. They are of the same sign, making its graph, if it exists, an ellipse.

To determine whether this ellipse is horizontal or vertical, rewrite this equation in standard form

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

as follows:

Separate the and terms:

$16x^{2} + 12y^{2} -96x-48y =0$

$(16x^{2} -96x) + (12y^{2}-48y) =0$

Distribute out the quadratic coefficients:

$16 ( x^{2} -6x ) +12 ( y^{2}-4y) =0$

Complete the square within each quadratic expression by dividing each linear coefficient by 2 and squaring the quotient.

Since $\left (\frac{-6}{2} \right )^{2}= 9$ and $\left (\frac{-4}{2} \right )^{2}= 4$ , we get

$16 ( x^{2} -6x \textbf{+ 9} ) +12 ( y^{2}-4y\textbf{ + 4}) =0+ \textbf{?}+ \textbf{?}$

Balance this equation, adjusting for the distributed coefficients:

$16 ( x^{2} -6x + 9 ) +12 ( y^{2}-4y + 4 ) =16(9)+ 12(4)$

$16 ( x^{2} -6x + 9 ) +12 ( y^{2}-4y + 4 ) =192$

The perfect square trinomials are squares of binomials, by design; rewrite them as such:

$16 ( x-3 )^{2} +12 ( y -2 ) ^{2} =192$

Divide by 192:

$\frac{16 ( x-3 )^{2}}{192} +\frac{12 ( y -2 ) ^{2} }{192}=\frac{192}{192}$

$\frac{( x-3 )^{2}}{12} +\frac{( y -2 ) ^{2} }{16}=1$

The ellipse is now in standard form. $b^{2} > a^{2}$ , so the graph of the equation is a vertical ellipse

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Question

The graph of the equation

$16x^{2}+12y^{2}-96x+48y+384=0$

is an example of which conic section?

Answer

The quadratic coefficients in this general form of a conic equation are 16 and 12. They are of the same sign, making its graph, if it exists, an ellipse.

To determine whether this ellipse is horizontal or vertical, rewrite this equation in standard form

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

as follows:

Subtract 384 from both sides:

$16x^{2}+12y^{2}-96x+48y+384=0$

$16x^{2}+12y^{2}-96x+48y= -384$

Separate the and terms and group them:

$16x^{2} + 12y^{2} -96x-48y =-384$

$(16x^{2} -96x) + (12y^{2}-48y) =-384$

Distribute out the quadratic coefficients:

$16 ( x^{2} -6x ) +12 ( y^{2}-4y) =-384$

Complete the square within each quadratic expression by dividing each linear coefficient by 2 and squaring the quotient.

Since $\left (\frac{-6}{2} \right )^{2}= 9$ and $\left (\frac{-4}{2} \right )^{2}= 4$ , we get

$16 ( x^{2} -6x \textbf{+ 9} ) +12 ( y^{2}-4y\textbf{ + 4}) =-384+ \textbf{?}+ \textbf{?}$

Balance this equation, adjusting for the distributed coefficients:

$16 ( x^{2} -6x + 9 ) +12 ( y^{2}-4y + 4 ) =-384+ 16(9)+ 12(4)$

$16 ( x^{2} -6x + 9 ) +12 ( y^{2}-4y + 4 ) =-192$

The perfect square trinomials are squares of binomials, by design; rewrite them as such:

$16 ( x-3 )^{2} +12 ( y -2 ) ^{2} =-192$

Divide by :

$\frac{16 ( x-3 )^{2}}{-192} +\frac{12 ( y -2 ) ^{2} }{-192}=\frac{-192}{-192}$

$\frac{( x-3 )^{2}}{-12} +\frac{( y -2 ) ^{2} }{-16}=1$

Recall that the standard form of an ellipse is

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

This requires both denominators to be positive. In the standard form of the given equation, they are not. Therefore, the equation has no real ordered pairs as solutions, and it does not have a graph on the coordinate plane.

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Question

Give the foci of the ellipse of the equation

$\frac{x^{2}}{46}+ \frac{y^{2}}{19} = 1$

Round your coordinates to the nearest tenth, if applicable.

Answer

The equation of the ellipse is given in the standard form

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

This ellipse has its center at the origin . Also, since $a^{2} >b^{2}$ , it follows that the ellipse is horizontal. The foci are therefore along the horizontal axis of the ellipse; their coordinates are , where

$c = \sqrt{a^{2}-b^{2}}$

Substituting 46 and 19 for $a^{2}$ and $b^{2}$ , respectively,

$c = \sqrt{46-19} = \sqrt{27} \approx 5.2$ .

The foci are at the points and .

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Question

Give the eccentricity of the ellipse of the equation

$\frac{x^{2}}{25}+ \frac{y^{2}}{169} = 1$

Answer

This ellipse is in standard form

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

where $b^{2} > a^{2}$ . This is a vertical ellipse, whose foci are

$c= \sqrt{b^{2}-a^{2}}$

units from its center in a vertical direction.

The eccentricity of this ellipse can be calculated by taking the ratio $\frac{c}{b}$ , or, equivalently, $\frac{\sqrt{b^{2}-a^{2}}}{b}$ . Set $b^{2} = 169$ - making - and $a^{2} = 25$ . The eccentricity is calculated to be

$\frac{\sqrt{169-25}}{13} =\frac{ \sqrt{144}}{13} = \frac{12}{13}$ .

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Question

Give the foci of the ellipse of the equation

$\frac{(x- 2)^{2}}{64} + \frac{(y-6)^{2}}{49} = 1$ .

Round your coordinates to the nearest tenth, if applicable.

Answer

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

is the standard form of an ellipse with center . Also, since in the given equation, $a^{2} = 64$ and $b^{2} = 49$ - that is, $a^{2}> b^{2}$ , the ellipse is horizontal.

The foci of a horizontal ellipse are located at

$(h \pm c ,k)$ ,

where

$c= \sqrt{a^{2}-b^{2}} = \sqrt{64-49} = \sqrt{15} \approx 3.9$

Setting $h = 2, k = 6, c \approx 3.9$ , the foci are at

$(2 \pm 3.9, 6)$ , or

and .

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Question

Using the information below, determine the equation of the hyperbola.

Foci: and

Eccentricity: $\frac{3}{2}$

Answer

General Information for Hyperbola:

Equation for horizontal transverse hyperbola:

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

Distance between foci =

Distance between vertices =

Eccentricity =

$\small a^{2}+b^{2}=c^{2}$

Center: (h, k)

First determine the value of c. Since we know the distance between the two foci is 12, we can set that equal to .

$\small c= 6$

Next, use the eccentricity equation and the value of the eccentricity provided in the question to determine the value of a.

Eccentricity = $\small \small c/a=3/2$

$\small \frac{c}{a}=\frac{3}{2}=\frac{6}{a}$

$\small 12=3a$

$\small a=4$

$\small \small a^{2}=16$

Determine the value of $\small b^{2}$

$\small a^{2}+b^{2}=c^{2}$

$\small \small 4^{2}+b^{2}=6^{2}$

$\small \small b^{2}=20$

Determine the center point to identify the values of h and k. Since the y coordinate of the foci are 4, the center point will be on the same line. Hence, $\small k = 4$ .

Since center point is equal distance from both foci, and we know that the distance between the foci is 12, we can conclude that $\small h=1$

Center point: $\small \small (h, k)= (1,4)$

Thus, the equation of the hyperbola is:

$\small \small \frac{(x-1)^{2}}{16} - \frac{(y-4)^{2}}{20}=1$

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Question

Using the information below, determine the equation of the hyperbola.

Foci: and

Eccentricity:

Answer

General Information for Hyperbola:

Equation for horizontal transverse hyperbola:

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

Distance between foci =

Distance between vertices =

Eccentricity =

$\small a^{2}+b^{2}=c^{2}$

Center: (h, k)

First determine the value of c. Since we know the distance between the two foci is 8, we can set that equal to .

$\small 8=2c$

$\small \small c= 4$

Next, use the eccentricity equation and the value of the eccentricity provided in the question to determine the value of a.

Eccentricity = $\small \small \small \small c/a=2$

$\small \small \frac{c}{a}=\frac{2}{1}=\frac{4}{a}$

$\small \small 4=2a$

$\small \small a=2$

$\small \small \small a^{2}=4$

Determine the value of $\small b^{2}$

$\small a^{2}+b^{2}=c^{2}$

$\small \small \small 2^{2}+b^{2}=4^{2}$

$\small \small \small b^{2}=12$

Determine the center point to identify the values of h and k. Since the y coordinate of the foci are 8, the center point will be on the same line. Hence, $\small \small k = 8$ .

Since center point is equal distance from both foci, and we know that the distance between the foci is 8, we can conclude that $\small \small h=5$

Center point: $\small \small \small (h, k)= (5, 8 )$

Thus, the equation of the hyperbola is:

$\small \small \small \small \small \small \frac{(x-5)^{2}}{4} - \frac{(y-8)^{2}}{12}=1$

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Question

Find the center and the vertices of the following hyperbola:

$\frac{(x+9)^2}{25}-\frac{(y-3)^2}{16}=1$

Answer

In order to find the center and the vertices of the hyperbola given in the problem, we must examine the standard form of a hyperbola:

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

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

The point (h,k) gives the center of the hyperbola. We can see that the equation in this problem resembles the second option for standard form above, so right away we can see the center is at:

In the first option, where the x term is in front of the y term, the hyperbola opens left and right. In the second option, where the y term is in front of the x term, the hyperbola opens up and down. In either case, the distance tells how far above and below or to the left and right of the center the vertices of the hyperbola are. Our equation is in the first form, where the x term is first, so the hyperbola opens left and right, which means the vertices are a distance to the left and right of the center. We can now calculate by identifying it in our equation, and then go 3 units to the left and right of our center to find the following vertices:

$\frac{(x+9)^2}{25}-\frac{(y-3)^2}{16}=1$

$a^2=25\rightarrow a=5$

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Question

Find the equations of the asymptotes for the hyperbola with the following equation:

Answer

For a hyperbola with its foci on the -axis, like the one given in the equation, recall the standard form of the equation:

$\frac{(y-k)^2}{b^2}-\frac{(x-h)^2}{a^2}=1$ , where is the center of the hyperbola.

Start by putting the given equation into the standard form of the equation of a hyperbola.

Group the terms together and terms together.

Factor out from the terms and from the terms.

Complete the squares. Remember to add the amount amount to both sides of the equation!

Add to both sides of the equation:

Divide both sides by .

$\frac{y^2-4y+4}{16}-\frac{x^2+2x+1}{36}=1$

Factor the two terms to get the standard form of the equation of a hyperbola.

$\frac{(y-2)^2}{16}-\frac{(x+1)^2}{36}=1$

The slopes of this hyperbola are given by the following:

$m=\pm\frac{b}{a}$

For the hyperbola in question, and .

Thus, the slopes for its asymptotes are $\pm\frac{4}{6}=\pm\frac{2}{3}$ .

Now, plug in the center of the hyperbola, into the point-slope form of the equation of a line to get the equations of the asymptotes.

For the first equation,

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

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

$y=\frac{2}{3}x+\frac{8}{3}$

For the second equation,

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

$y-2=-\frac{2}{3}x-\frac{2}{3}$

$y=-\frac{2}{3}x+\frac{4}{3}$

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Question

Find the foci of the hyperbola with the following equation:

$\frac{x^2}{27}-\frac{y^2}{9}=1$

Answer

Recall that the standard formula of a hyperbola can come in two forms:

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

$\frac{(y-k)^2}{b^2}-\frac{(x-h)^2}{a^2}-=1$ , where the centers of both hyperbolas are .

When the term with is first, that means the foci will lie on a horizontal transverse axis.

When the term with is first, that means the foci will lie on a vertical transverse axis.

To find the foci, we use the following:

$c=\sqrt{a^2+b^2}$

For a hyperbola with a horizontal transverse access, the foci will be located at and .

For a hyperbola with a vertical transverse access, the foci will be located at and .

For the given hypebola in the question, the transverse axis is horizontal and its center is located at .

Next, find .

$c=\sqrt{27+9}=\sqrt{36}=6$

The foci are then located at and .

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Question

Find the foci of a hyperbola with the following equation:

$\frac{x^2}{36}-\frac{y^2}{64}=1$

Answer

Recall that the standard formula of a hyperbola can come in two forms:

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

$\frac{(y-k)^2}{b^2}-\frac{(x-h)^2}{a^2}-=1$ , where the centers of both hyperbolas are .

When the term with is first, that means the foci will lie on a horizontal transverse axis.

When the term with is first, that means the foci will lie on a vertical transverse axis.

To find the foci, we use the following:

$c=\sqrt{a^2+b^2}$

For a hyperbola with a horizontal transverse access, the foci will be located at and .

For a hyperbola with a vertical transverse access, the foci will be located at and .

For the given hypebola in the question, the transverse axis is horizontal and its center is located at .

Next, find .

$c=\sqrt{64+36}=\sqrt{100}=10$

The foci are then located at and .

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Question

Find the foci of the hyperbola with the following equation:

$\frac{(x+2)^2}{16}-\frac{(y-2)^2}{9}=1$

Answer

Recall that the standard formula of a hyperbola can come in two forms:

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

$\frac{(y-k)^2}{b^2}-\frac{(x-h)^2}{a^2}-=1$ , where the centers of both hyperbolas are .

When the term with is first, that means the foci will lie on a horizontal transverse axis.

When the term with is first, that means the foci will lie on a vertical transverse axis.

To find the foci, we use the following:

$c=\sqrt{a^2+b^2}$

For a hyperbola with a horizontal transverse access, the foci will be located at and .

For a hyperbola with a vertical transverse access, the foci will be located at and .

For the given hypebola in the question, the transverse axis is horizontal and its center is located at .

Next, find .

$c=\sqrt{16+9}=\sqrt{25}=5$

The foci are then located at and .

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Question

Find the foci of the hyperbola with the following equation:

Answer

Recall that the standard formula of a hyperbola can come in two forms:

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

$\frac{(y-k)^2}{b^2}-\frac{(x-h)^2}{a^2}-=1$ , where the centers of both hyperbolas are .

First, put the given equation in the standard form of the equation of a hyperbola.

Group the terms together and the terms together.

Next, factor out from the terms and from the terms.

From here, complete the squares. Remember to add the same amount to both sides of the equation!

Subtract from both sides.

Divide both sides by .

$\frac{x^2-8x+16}{14}-\frac{y^2-4y+4}{50}=1$

Factor both terms to get the standard form for the equation of a hyperbola.

$\frac{(x-4)^2}{14}-\frac{(y-2)^2}{50}=1$

When the term with is first, that means the foci will lie on a horizontal transverse axis.

When the term with is first, that means the foci will lie on a vertical transverse axis.

To find the foci, we use the following:

$c=\sqrt{a^2+b^2}$

For a hyperbola with a horizontal transverse access, the foci will be located at and .

For a hyperbola with a vertical transverse access, the foci will be located at and .

For the given hypebola in the question, the transverse axis is horizontal and its center is located at .

Next, find .

$c=\sqrt{14+50}=\sqrt{64}=8$

The foci are then located at and .

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