Answer:
1) x-intercepts = ±5
y-intercepts = ±3
foci = ±4
2) x-intercepts = ±4
y-intercepts = ±2
foci = ±2√3
3) e = 0.66
Step-by-step explanation:
The general equation of an ellipse with center (h, k) is:
[tex]\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1[/tex]
If a > b the ellipse is horizontal, 2a is the major axis, and 2b is the minor axis.
If b > a the ellipse is vertical, 2b is the major axis, and 2a is the minor axis.
[tex]\dotfill[/tex]
Given equation of an ellipse:
[tex]\dfrac{x^2}{25}+\dfrac{y^2}{9}=1[/tex]
The x-intercepts are the x-coordinates of the points at which the curve intersects the x-axis, so when y = 0. To find the x-intercepts of the ellipse, substitute y = 0 into the equation and solve for x:
[tex]\dfrac{x^2}{25}+\dfrac{0^2}{9}=1\\\\\\\dfrac{x^2}{25}+0=1\\\\\\\dfrac{x^2}{25}=1\\\\\\x^2=25\\\\\\\sqrt{x^2}=\sqrt{25}\\\\\\x=\pm 5[/tex]
Therefore, the x-intercepts are ±5.
The y-intercepts are the y-coordinates of the points at which the curve intersects the y-axis, so when x = 0. To find the y-intercepts of the ellipse, substitute x = 0 into the equation and solve for y:
[tex]\dfrac{0^2}{25}+\dfrac{y^2}{9}=1\\\\\\0+\dfrac{y^2}{9}=1\\\\\\\dfrac{y^2}{9}=1\\\\\\y^2=9\\\\\\\sqrt{y^2}=\sqrt{9}\\\\\\y=\pm3[/tex]
Therefore, the y-intercepts are ±3.
As the denominator of the x² term is greater than the denominator of the y² term, the ellipse is horizontal, which means that the major axis (longest diameter of the ellipse) is parallel to the x-axis.
The foci of an ellipse lie on its major axis. The formula for the focus of a horizontal ellipse is (h±c, k) where (h, k) is the center of the ellipse and c² = a² - b².
[tex]c^2=25-9\\\\c^2=16\\\\\sqrt{c^2}=\sqrt{16}\\\\c=\pm 4[/tex]
In this case, the center of the ellipse is the origin, so h = 0 and k = 0. Therefore:
[tex]\textsf{Foci}=(0\pm 4, 0)\\\\\textsf{Foci}=(\pm 4, 0)[/tex]
So, the foci are ±4.
[tex]\dotfill[/tex]
Given equation of an ellipse:
[tex]16y^2=64-4x^2[/tex]
The x-intercepts are the x-coordinates of the points at which the curve intersects the x-axis, so when y = 0. To find the x-intercepts of the ellipse, substitute y = 0 into the equation and solve for x:
[tex]16(0)^2=64-4x^2\\\\0=64-4x^2\\\\4x^2-64=0\\\\4(x^2-16)=0\\\\x^2-16=0\\\\x^2=16\\\\\sqrt{x^2}=\sqrt{16}\\\\x=\pm 4[/tex]
Therefore, the x-intercepts are ±4.
The y-intercepts are the y-coordinates of the points at which the curve intersects the y-axis, so when x = 0. To find the y-intercepts of the ellipse, substitute x = 0 into the equation and solve for y:
[tex]16y^2=64-4(0)\\\\16y^2=64-0\\\\16y^2=64\\\\y^2=4\\\\\sqrt{y^2}=\sqrt{4}\\\\y=\pm2[/tex]
Therefore, the y-intercepts are ±2.
Rearrange the equation into the general form of an ellipse:
[tex]16y^2=64-4x^2\\\\\\4x^2+16y^2=64\\\\\\4(x^2+4y^2)=4(16)\\\\\\x^2+4y^2=16\\\\\\\dfrac{x^2}{16}+\dfrac{4y^2}{16}=1\\\\\\\dfrac{x^2}{16}+\dfrac{y^2}{4}=1[/tex]
As the denominator of the x² term is greater than the denominator of the y² term, the ellipse is horizontal, which means that the major axis (longest diameter of the ellipse) is parallel to the x-axis.
The foci of an ellipse lie on its major axis. The formula for the focus of a horizontal ellipse is (h±c, k) where (h, k) is the center of the ellipse and c² = a² - b².
[tex]c^2=16-4\\\\c^2=12\\\\\sqrt{c^2}=\sqrt{12}\\\\c=\pm 2\sqrt{3}[/tex]
In this case, the center of the ellipse is the origin, so h = 0 and k = 0. Therefore:
[tex]\textsf{Foci}=(0\pm 2\sqrt{3}, 0)\\\\\textsf{Foci}=(\pm 2\sqrt{3}, 0)[/tex]
So, the foci are ±2√3.
[tex]\dotfill[/tex]
Given equation of an ellipse:
[tex]\dfrac{(x+2)^2}{16}+\dfrac{(y+1)^2}{9}=1[/tex]
As the denominator of the x² term is greater than the denominator of the y² term, the ellipse is horizontal, which means that the major axis (longest diameter of the ellipse) is parallel to the x-axis.
The formula for the eccentricity (e) of a horizontal ellipse is given by:
[tex]e = \sqrt{1 - \dfrac{b^2}{a^2}}[/tex]
where:
Key points about eccentricity:
In this case, a² = 16 and b² = 9. Therefore:
[tex]e = \sqrt{1 - \dfrac{9}{16}}\\\\\\e = \sqrt{\dfrac{16}{16} - \dfrac{9}{16}}\\\\\\e = \sqrt{\dfrac{7}{16}}\\\\\\e=0.661437827766...\\\\\\e=0.66\; \sf (nearest\;hundreth)[/tex]
So, the value of e is 0.66.