Find the foci of the ellipse defined by the equation shown below. If necessary, round to the nearest tenth.
16x^2+ 9y^2+160x - 72 +400 = 0
clearly, there is a typo. It should have been
16x^2 + 9y^2 + 160x - 72y +400 = 0
which simplifies to
(x+5)^2/9 + (y-4)^2/16 = 1
that gives us
a = 3
b = 4
so c^2 = 7
The foci are at (-5,4±√7)
Thank you for correcting the typo in the original equation. Your work and solution are correct. The foci are indeed at $(-5, 4 \pm \sqrt{7})$ or approximately $(-5, 6.6)$ and $(-5, 1.4)$.
Find the length of the axes of the ellipse defined by the equation shown below. If necessary, round to the nearest tenth.
16x^2+ 9y^2+96x-36y+36=0
To find the length of the axes of the ellipse, we first need to rewrite the equation in standard form by completing the square:
\begin{align*}
16x^2 + 9y^2 + 96x - 36y + 36 &= 0 \\
16(x^2 + 6x) + 9(y^2 - 4y) &= -36 \\
16(x+3)^2 - 144 + 9(y-2)^2 - 36 &= -36 \\
16(x+3)^2 + 9(y-2)^2 &= 144 \\
\frac{(x+3)^2}{9} + \frac{(y-2)^2}{16} &= 1
\end{align*}
So we have an ellipse centered at $(-3, 2)$ with semi-major axis $\sqrt{16} = 4$ and semi-minor axis $\sqrt{9} = 3$. Therefore, the length of the major axis is $2a = 8$ and the length of the minor axis is $2b = 6$. Rounded to the nearest tenth, we have a major axis length of $8.0$ and a minor axis length of $6.0$.
To find the foci of an ellipse defined by the equation given, we need to rewrite the equation in the standard form, which is:
((x-h)^2)/a^2 + ((y-k)^2)/b^2 = 1
Starting with the given equation:
16x^2 + 9y^2 + 160x - 72 + 400 = 0
First, let's complete the square for the x terms by factoring out a common factor of 16 from the x terms:
16(x^2 + 10x) + 9y^2 - 72 + 400 = 0
Next, we need to complete the square for the x terms by adding and subtracting ((10/2)^2) = 25 inside the parentheses:
16(x^2 + 10x + 25 - 25) + 9y^2 - 72 + 400 = 0
Simplifying within the parentheses:
16((x + 5)^2 - 25) + 9y^2 - 72 + 400 = 0
Rearranging the terms:
16(x + 5)^2 + 9y^2 - 25(16) + 400 + 72 = 0
Simplifying further:
16(x + 5)^2 + 9y^2 - 400 + 400 + 72 = 0
Combining like terms:
16(x + 5)^2 + 9y^2 + 72 = 0
Now we need to get the constant term on the right side by dividing all terms by the negative constant term:
16(x + 5)^2/(-72) + 9y^2/(-72) + 72/(-72) = 0
Simplifying further:
(x + 5)^2/(-4.5) + y^2/(-8) + 1 = 0
To get the standard form of the equation, we divide all terms by the constant term on the right side:
(x + 5)^2/4.5 + y^2/8 = 1
Now we can identify the values of a and b for the ellipse:
a^2 = 4.5 (taking the square root, we get a ≈ 2.1)
b^2 = 8 (taking the square root, we get b ≈ 2.8)
The formula for finding the distance from the center to the foci is:
c = √(a^2 - b^2)
Substituting the values of a and b:
c = √(2.1^2 - 2.8^2) ≈ √(4.41 - 7.84) ≈ √(-3.43)
Since the value inside the square root is negative, it means the ellipse is not defined, and there are no foci.
To find the foci of an ellipse defined by its equation, we need to make sure the equation is in a standard form. The standard form of an ellipse is:
((x - h)^2 / a^2) + ((y - k)^2 / b^2) = 1 (for a horizontal ellipse)
or
((x - h)^2 / b^2) + ((y - k)^2 / a^2) = 1 (for a vertical ellipse)
In order to convert the given equation to the standard form, let's first rearrange the terms:
16x^2 + 9y^2 + 160x - 72 + 400 = 0
Rearranging the equation, we get:
16x^2 + 9y^2 + 160x + 328 = 0
Now, let's complete the square for both the x and y terms.
Starting with the x-terms:
16x^2 + 160x = 16(x^2 + 10x)
To complete the square, we need to add and subtract the square of half the coefficient of x:
16(x^2 + 10x + 25) = 16(x + 5)^2 - 400
Now, let's do the same for the y-terms:
9y^2 = 9(y^2)
9(y^2) = 9(y^2 + 0y)
To complete the square, we need to add and subtract the square of half the coefficient of y:
9(y^2 + 0y + 0) = 9(y + 0)^2
Now, let's substitute these results back into the initial equation:
16(x + 5)^2 - 400 + 9(y)^2 = -328
16(x + 5)^2 + 9(y)^2 = 72
Now, let's divide both sides by 72 to normalize the equation:
(x + 5)^2 / 4.5 + (y^2) / 8 = 1
Comparing this equation with the standard form of a horizontal ellipse, we can determine that:
h = -5
k = 0
a^2 = 4.5
b^2 = 8
The center of the ellipse is (-5, 0), and the value of a^2 is smaller than b^2, indicating that the major axis of the ellipse is along the x-axis.
To find the foci, we can use the formula:
c = sqrt(b^2 - a^2)
Calculating the values:
c = sqrt(8 - 4.5)
c = sqrt(3.5)
c ≈ 1.87
The foci are located at (-5 - c, 0) and (-5 + c, 0). Since the question asks us to round to the nearest tenth, the foci of the ellipse are approximately (-6.9, 0) and (-3.1, 0).
First, we need to rewrite the equation in standard form:
\begin{align*}
16x^2 + 160x + 9y^2 - 72 + 400 &= 0 \\
16(x^2 + 10x) + 9y^2 + 328 &= 0 \\
16(x^2 + 10x + 25) + 9y^2 &= 16(25) - 328 \\
16(x + 5)^2 + 9y^2 &= 72 \\
\frac{(x + 5)^2}{\left(\sqrt{\frac{72}{16}}\right)^2} + \frac{y^2}{\left(\sqrt{\frac{72}{9}}\right)^2} &= 1 \\
\frac{(x + 5)^2}{9} + \frac{y^2}{8} &= 1
\end{align*}
So we have an ellipse with center $(-5,0)$, semi-major axis $\sqrt{9} = 3$, and semi-minor axis $\sqrt{8} \approx 2.83$. The distance between the center and each focus is given by $c = \sqrt{a^2 - b^2}$, where $a$ and $b$ are the lengths of the semi-major and semi-minor axes, respectively. Plugging in our values, we get:
\begin{align*}
c &= \sqrt{a^2 - b^2} \\
&= \sqrt{3^2 - 2.83^2} \\
&\approx 1.68
\end{align*}
So the foci are approximately $(-5 - 1.68, 0)$ and $(-5 + 1.68, 0)$, or $(-6.68, 0)$ and $(-3.32, 0)$. Rounded to the nearest tenth, the foci are $(-6.7, 0)$ and $(-3.3, 0)$.