Understand features of hyperbolas and ellipses - Pre-Calculus

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Question

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

$\frac{(x-2)^2}{100}-\frac{(y-6)^2}{36}=1$

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Answer

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

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

The slopes of the asymptotes for 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 $m=\pm\frac{6}{10}=\pm\frac{3}{5}$ .

Now, use the point-slope form of a line in addition to the center of the hyperbola to find the equations of the asymptotes.

The center is at .

The first equation of the asymptote would be the following:

$y-6=\frac{3}{5}(x-2)$

$y-6=\frac{3}{5}x-\frac{6}{5}$

$y=\frac{3}{5}x+\frac{24}{5}$

The second equation of the asymptote would be the following:

$y-6=-\frac{3}{5}(x-2)$

$y-6=-\frac{3}{5}x+\frac{6}{5}$

$y=-\frac{3}{5}+\frac{36}{5}$

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Question

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

$\frac{(x-5)^2}{24}-\frac{(y+3)^2}{25}=1$

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Answer

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

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

The slopes of the asymptotes for this hyperbola are given by the following:

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

For the hyperbola in question, $a=\sqrt{24}=2\sqrt6$ and .

Thus, the slopes for its asymptotes are $m=\pm\frac{5}{2\sqrt6}=\pm\frac{5\sqrt6}{12}$ .

Now, use the point-slope form of a line in addition to the center of the hyperbola to find the equations of the asymptotes.

The center is at .

To find the equations of the asymptotes, use the point-slope form of a line.

$y+3=\pm\frac{5\sqrt6}{12}(x-5)$

$y=\frac{5\sqrt6}{12}(x-5)-3$

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Question

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

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Answer

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

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

Subtract from both sides of the equation:

Divide both sides by .

$\frac{x^2-18x+81}{4}-\frac{y^2+2y+1}{9}=1$

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

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

The slopes of the asymptotes for 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 $m=\pm\frac{3}{2}$ .

Plug in the center of the hyperbola into the point-slope form of a line to find the equations of the asymptotes. The center of the hyperbola is .

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

Now, simplify each equation. For the first equation,

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

$y+1=\frac{3}{2}x-\frac{27}{2}$

$y=\frac{3}{2}x-\frac{29}{2}$

For the second equation,

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

$y+1=-\frac{3}{2}x+\frac{27}{2}$

$y=-\frac{3}{2}x-\frac{25}{2}$

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Question

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

$\frac{(x-12)^2}{144}-\frac{(y+\sqrt3)^2}{169}=1$

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Answer

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

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

The slopes of the asymptotes for 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 $m=\pm\frac{13}{12}$

Now, plug in the coordinates for the center of the hyperbola $(12, \sqrt3)$ into the point-slope form of the line to find the equations of the asymptotes.

$y-\sqrt3=\pm\frac{13}{12}(x-12)$

$y=\pm\frac{13}{12}(x-12)+\sqrt3$

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Question

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

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Answer

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

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

$\frac{x^2+6x+9}{25}-\frac{y^2-22y+121}{4}=1$

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

$\frac{(x+3)^2}{25}-\frac{(y-11)^2}{4}=1$

The slopes of the asymptotes for 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 $\pm\frac{2}{5}$ .

Now, use the center of the hyperbola, , to plug into the point-slope form of a line to find the equations of the asymptotes.

The first asymptote has the following equation:

$y-11=\frac{2}{5}(x+3)$

$y-11=\frac{2}{5}x+\frac{6}{5}$

$y=\frac{2}{5}x+\frac{61}{5}$

The second asymptote has the following equation:

$y-11=-\frac{2}{5}(x+3)$

$y-11=-\frac{2}{5}x-\frac{6}{5}$

$y=-\frac{2}{5}x+\frac{49}{5}$

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Question

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

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

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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.

The slopes of the asymptotes for this hyperbola are given by the following:

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

For the hyperbola in question, and .

Since the center is , the equations for its asymptotes are $y=\pm\frac{3}{5}x$ .

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Question

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

$\frac{(y-9)^2}{169}-\frac{(x-2)^2}{225}=1$

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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.

The slopes of the asymptotes for 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 $\pm\frac{13}{15}$ .

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

For the first asymptote,

$y-9=\frac{13}{15}(x-2)$

$y-9=\frac{13}{15}x-\frac{26}{15}$

$y=\frac{13}{15}x+\frac{109}{15}$

For the second asymptote,

$y-9=-\frac{13}{15}(x-2)$

$y-9=-\frac{13}{15}x+\frac{26}{15}$

$y=-\frac{13}{15}x+\frac{161}{15}$

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Question

Find the endpoints of the minor axis of the ellipse with the following equation:

$\frac{(x-5)^2}{25}+\frac{(y-1)^2}{36}=1$

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Answer

Recall 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.

When , the minor axis is horizontal. In this case, and are the endpoints of the minor axis.

When , and are the endpoints of the vertical minor axis.

For the ellipse in question, is the center. In addition, and . Since , the minor axis is horizontal and the endpoints are and .

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Question

Find the endpoints of the minor axis of the ellipse with the following equation:

$\frac{(x-1)^2}{64}+\frac{(y-3)^2}{25}=1$

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Answer

Recall 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.

When , the minor axis is horizontal. In this case, and are the endpoints of the minor axis.

When , and are the endpoints of the vertical minor axis.

For the ellipse in question, is the center. In addition, and . Since , the minor axis is vertical and the endpoints are and .

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Question

Find the endpoints of the minor axis of the ellipse with the following equation:

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Answer

Recall 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.

Start by putting the equation in the standard form as shown above.

Group the terms and terms together.

$25x^2-50x+64y^2-384y-999=0\textup{}$

Factor out from the terms and from the terms.

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

Add to both sides.

Divide by on both sides.

$\frac{x^2-2x+1}{64}+\frac{y^2-6y+9}{25}=1$

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

$\frac{(x-1)^2}{64}+\frac{(y-3)^2}{25}=1$

When , the minor axis is horizontal. In this case, and are the endpoints of the minor axis.

When , and are the endpoints of the vertical minor axis.

For the ellipse in question, is the center. In addition, and . Since , the minor axis is vertical and the endpoints are and .

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Question

Find the endpoints of the minor axis of the ellipse with the following equation:

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Answer

Recall 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.

Start by putting the equation in the standard form as shown above.

Group the terms and terms together.

Factor out from the terms and from the terms.

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

Subtract from both sides.

Divide by on both sides.

$\frac{x^2+8x+16}{20}+\frac{y^2-6x+9}{25}=1$

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

$\frac{(x+4)^2}{20}+\frac{(y-3)^2}{25}=1$

When , the minor axis is horizontal. In this case, and are the endpoints of the minor axis.

When , and are the endpoints of the vertical minor axis.

For the ellipse in question, is the center. In addition, $a=\sqrt{20}=2\sqrt5$ and . Since , the minor axis is horizontal and the endpoints are $(-4+2\sqrt5, 3)$ and $(-4-2\sqrt5, 3)$ .

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Question

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

$\frac{x^2}{36}-\frac{y^2}{49}=1$

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Answer

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

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

The equations of the asymptotes for this hyperbola are given by the following equations:

$y=\pm\frac{b}{a}x$

For the hyperbola in question, and .

Thus, the equations for its asymptotes are $y=\pm\frac{7}{6}x$

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Question

Find the endpoints of the major axis of the ellipse with the following equation:

$\frac{(x-2)^2}{225}+\frac{(y+11)^2}{256}=1$

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Answer

Recall 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.

When , the major axis is horizontal. In this case, and are the endpoints of the major axis.

When , and are the endpoints of the major axis.

For the ellipse in question, is the center. In addition, and . Since , the major axis is vertical and the endpoints are and .

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Question

Find the endpoints of the major axis of the ellipse with the following equation:

$\frac{(x+5)^2}{169}+\frac{(y-3)^2}{144}=1$

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Answer

Recall 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.

When , the major axis is horizontal. In this case, and are the endpoints of the major axis.

When , and are the endpoints of the major axis.

For the ellipse in question, is the center. In addition, and . Since , the major axis is horizontal and the endpoints are and

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Question

Find the endpoints of the major axis of the ellipse with the following equation:

Tap to reveal answer

Answer

Recall 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.

Start by putting the equation in the standard form as shown above.

Group the terms and terms together.

Factor out from the terms and from the terms.

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

Add to both sides.

Divide by on both sides.

$\frac{x^2-8x+16}{49}+\frac{y^2-6y+9}{25}=1$

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

$\frac{(x-4)^2}{49}+\frac{(y-3)^2}{25}=1$

Recall that when , the major axis is horizontal. In this case, and are the endpoints of the major axis.

When , and are the endpoints of the major axis.

For the ellipse in question, is the center. In addition, and . Since , the major axis is horizontal and the endpoints are and .

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Question

Find the endpoints of the major axis of the ellipse with the following equation:

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Answer

Recall 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.

Start by putting the equation in the standard form as shown above.

Group the terms and terms together.

Factor out from the terms and from the terms.

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

Subtract from both sides.

Divide by on both sides.

$\frac{x^2-12x+36}{18}+\frac{y^2-20x+100}{9}=1$

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

$\frac{(x-6)^2}{18}+\frac{(y-10)^2}{9}=1$

Recall that when , the major axis is horizontal. In this case, and are the endpoints of the major axis.

When , and are the endpoints of the major axis.

For the ellipse in question, is the center. In addition, $a=\sqrt{18}=3\sqrt2$ and . Since , the major axis is horizontal and the endpoints are $(6+3\sqrt2, 10)$ and $(6-3\sqrt2, 10)$ .

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Question

Find the endpoints of the minor axis of the ellipse with the following equation:

$\frac{x^2}{100}+\frac{y^2}{169}=1$

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Answer

Recall 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.

When , the minor axis is horizontal. In this case, and are the endpoints of the minor axis.

When , and are the endpoints of the vertical minor axis.

For the ellipse in question, is the center. In addition, and . Since , the minor axis is horizontal and the endpoints are and

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Question

Find the endpoints of the minor axis of the ellipse with the following equation:

$\frac{x^2}{225}+\frac{y^2}{400}=1$

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Answer

Recall 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.

When , the minor axis is horizontal. In this case, and are the endpoints of the minor axis.

When , and are the endpoints of the vertical minor axis.

For the ellipse in question, is the center. In addition, and . Since , the minor axis is horizontal and the endpoints are and

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Question

Consider the following equation:

Classify the equation by its form when graphed.

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Answer

Start with the equation

It would be helpful if we "complete the square" for both the x and y polynomials.

So by adding 9 and 900 we transform the equation into

And then by dividing 900, we see that the outcome is:

$\frac{(x-1)^2}{100} + \frac{(y+3)^2}{9} = 1$

The standard equation for an ellipse is

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

With a center at

Therefore this is the equation of an ellipse, with center

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Question

Express the following equation for an ellipse in standard form:

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Answer

Remember that the equation for an ellipse in standard form looks like the following:

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

Where the point (h,k) gives the center of the ellipse, a is half the length of its axis in the x direction, and b is half the length of its axis in the y direction. We can see that this form has a 1 on the right side of the equation, so let's start by dividing both sides of our equation by 36 to get a 1 on the right side:

$\frac{4(x-4)^2+9(x-7)^2}{36}=\frac{36}{36}$

$\frac{4(x-4)^2}{36}+\frac{9(x-7)^2}{36}=1$

Now we can simplify the fractions on the right side of the equation, which gives us the equation for our ellipse in standard form:

$\frac{(x-4)^2}{9}+\frac{(y-7)^2}{4}=1$

This ellipse would have its center at (4,7), would be 6 units wide in the x direction, and 4 units wide in the y direction, because so , and so .

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