Answer :
To compare the graphs of the hyperbolas [tex]\(\frac{x^2}{3^2} - \frac{y^2}{4^2} = 1\)[/tex] and [tex]\(\frac{y^2}{3^2} - \frac{x^2}{4^2} = 1\)[/tex], we can analyze their properties step by step. We need to identify all valid statements and pick the one that holds true regarding the graphs.
1. Understanding the structure of the hyperbolas:
The first equation is:
[tex]\[ \frac{x^2}{3^2} - \frac{y^2}{4^2} = 1 \][/tex]
This represents a hyperbola centered at the origin with a horizontal transverse axis.
The second equation is:
[tex]\[ \frac{y^2}{3^2} - \frac{x^2}{4^2} = 1 \][/tex]
This represents a hyperbola centered at the origin with a vertical transverse axis.
2. Analyzing the given statements:
- Statement 1: The foci of both graphs are the same points.
For the hyperbola [tex]\(\frac{x^2}{3^2} - \frac{y^2}{4^2} = 1\)[/tex], the distance of the foci from the center is:
[tex]\[ c = \sqrt{a^2 + b^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5. \][/tex]
The foci are [tex]\((\pm 5, 0)\)[/tex].
For the hyperbola [tex]\(\frac{y^2}{3^2} - \frac{x^2}{4^2} = 1\)[/tex], the distance of the foci from the center is also:
[tex]\[ c = \sqrt{a^2 + b^2} = \sqrt{3^2 + 4^2} = 5. \][/tex]
The foci are [tex]\((0, \pm 5)\)[/tex].
The foci are not the same points since they have different coordinates.
- Statement 2: The lengths of both transverse axes are the same.
The length of the transverse axis is [tex]\(2a\)[/tex].
For [tex]\(\frac{x^2}{3^2} - \frac{y^2}{4^2} = 1\)[/tex], the length of the transverse axis is [tex]\(2 \times 3 = 6\)[/tex].
For [tex]\(\frac{y^2}{3^2} - \frac{x^2}{4^2} = 1\)[/tex], the length of the transverse axis is [tex]\(2 \times 3 = 6\)[/tex].
Both transverse axes are indeed the same length, but we need to check other statements as well.
- Statement 3: The directrices of [tex]\(\frac{x^2}{3^2} - \frac{y^2}{4^2} = 1\)[/tex] are horizontal while the directrices of [tex]\(\frac{y^2}{3^2} - \frac{x^2}{4^2} = 1\)[/tex] are vertical.
For [tex]\(\frac{x^2}{3^2} - \frac{y^2}{4^2} = 1\)[/tex]:
- The transverse axis is horizontal, so the directrices are vertical.
For [tex]\(\frac{y^2}{3^2} - \frac{x^2}{4^2} = 1\)[/tex]:
- The transverse axis is vertical, so the directrices are horizontal.
This statement is indeed correct.
- Statement 4: The vertices of [tex]\(\frac{x^2}{3^2} - \frac{y^2}{4^2} = 1\)[/tex] are on the y-axis while the vertices of [tex]\(\frac{y^2}{3^2} - \frac{x^2}{4^2} = 1\)[/tex] are on the x-axis.
For [tex]\(\frac{x^2}{3^2} - \frac{y^2}{4^2} = 1\)[/tex]:
- The vertices are at [tex]\((\pm 3, 0)\)[/tex], which are on the x-axis.
For [tex]\(\frac{y^2}{3^2} - \frac{x^2}{4^2} = 1\)[/tex]:
- The vertices are at [tex]\((0, \pm 3)\)[/tex], which are on the y-axis.
This statement is false as the axes mentioned are reversed.
Hence, the correct answer is:
```
The directrices of [tex]\(\frac{x^2}{3^2} - \frac{y^2}{4^2} = 1\)[/tex] are horizontal while the directrices of [tex]\(\frac{y^2}{3^2} - \frac{x^2}{4^2} = 1\)[/tex] are vertical.
```
This matches the third option of the query. The correct statement is:
- The directrices of [tex]\(\frac{x^2}{3^2} - \frac{y^2}{4^2} = 1\)[/tex] are horizontal while the directrices of [tex]\(\frac{y^2}{3^2} - \frac{x^2}{4^2} = 1\)[/tex] are vertical.
1. Understanding the structure of the hyperbolas:
The first equation is:
[tex]\[ \frac{x^2}{3^2} - \frac{y^2}{4^2} = 1 \][/tex]
This represents a hyperbola centered at the origin with a horizontal transverse axis.
The second equation is:
[tex]\[ \frac{y^2}{3^2} - \frac{x^2}{4^2} = 1 \][/tex]
This represents a hyperbola centered at the origin with a vertical transverse axis.
2. Analyzing the given statements:
- Statement 1: The foci of both graphs are the same points.
For the hyperbola [tex]\(\frac{x^2}{3^2} - \frac{y^2}{4^2} = 1\)[/tex], the distance of the foci from the center is:
[tex]\[ c = \sqrt{a^2 + b^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5. \][/tex]
The foci are [tex]\((\pm 5, 0)\)[/tex].
For the hyperbola [tex]\(\frac{y^2}{3^2} - \frac{x^2}{4^2} = 1\)[/tex], the distance of the foci from the center is also:
[tex]\[ c = \sqrt{a^2 + b^2} = \sqrt{3^2 + 4^2} = 5. \][/tex]
The foci are [tex]\((0, \pm 5)\)[/tex].
The foci are not the same points since they have different coordinates.
- Statement 2: The lengths of both transverse axes are the same.
The length of the transverse axis is [tex]\(2a\)[/tex].
For [tex]\(\frac{x^2}{3^2} - \frac{y^2}{4^2} = 1\)[/tex], the length of the transverse axis is [tex]\(2 \times 3 = 6\)[/tex].
For [tex]\(\frac{y^2}{3^2} - \frac{x^2}{4^2} = 1\)[/tex], the length of the transverse axis is [tex]\(2 \times 3 = 6\)[/tex].
Both transverse axes are indeed the same length, but we need to check other statements as well.
- Statement 3: The directrices of [tex]\(\frac{x^2}{3^2} - \frac{y^2}{4^2} = 1\)[/tex] are horizontal while the directrices of [tex]\(\frac{y^2}{3^2} - \frac{x^2}{4^2} = 1\)[/tex] are vertical.
For [tex]\(\frac{x^2}{3^2} - \frac{y^2}{4^2} = 1\)[/tex]:
- The transverse axis is horizontal, so the directrices are vertical.
For [tex]\(\frac{y^2}{3^2} - \frac{x^2}{4^2} = 1\)[/tex]:
- The transverse axis is vertical, so the directrices are horizontal.
This statement is indeed correct.
- Statement 4: The vertices of [tex]\(\frac{x^2}{3^2} - \frac{y^2}{4^2} = 1\)[/tex] are on the y-axis while the vertices of [tex]\(\frac{y^2}{3^2} - \frac{x^2}{4^2} = 1\)[/tex] are on the x-axis.
For [tex]\(\frac{x^2}{3^2} - \frac{y^2}{4^2} = 1\)[/tex]:
- The vertices are at [tex]\((\pm 3, 0)\)[/tex], which are on the x-axis.
For [tex]\(\frac{y^2}{3^2} - \frac{x^2}{4^2} = 1\)[/tex]:
- The vertices are at [tex]\((0, \pm 3)\)[/tex], which are on the y-axis.
This statement is false as the axes mentioned are reversed.
Hence, the correct answer is:
```
The directrices of [tex]\(\frac{x^2}{3^2} - \frac{y^2}{4^2} = 1\)[/tex] are horizontal while the directrices of [tex]\(\frac{y^2}{3^2} - \frac{x^2}{4^2} = 1\)[/tex] are vertical.
```
This matches the third option of the query. The correct statement is:
- The directrices of [tex]\(\frac{x^2}{3^2} - \frac{y^2}{4^2} = 1\)[/tex] are horizontal while the directrices of [tex]\(\frac{y^2}{3^2} - \frac{x^2}{4^2} = 1\)[/tex] are vertical.
