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 $(-3.3, 0)$.
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
5 answers
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)
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
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 $6.0$.
\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 $6.0$.