How do you find the center, foci and vertices of $\frac{{(x - 3)}^{2}}{\frac{25}{9}} + {(y - 8)}^{2} = 1$ $(x-3)^2/(25/9)+(y-8)^2=1$ ?

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How do you find the center, foci and vertices of $\frac{{(x - 3)}^{2}}{\frac{25}{9}} + {(y - 8)}^{2} = 1$ $(x-3)^2/(25/9)+(y-8)^2=1$ ?

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Answer 1 · 2016-12-28T23:36:47

Ellipse, center $(3, 8)$ $(3,8)$ ,Vertices $(\frac{4}{3}, 8)$ $(4/3,8)$ , $(\frac{14}{3}, 8)$ $(14/3,8)$ , $(3, 9)$ $(3,9)$ , $(3, 7)$ $(3,7)$ , foci $(3 \pm \frac{4}{3}, 8)$ $(3+-4/3,8)$

Explanation:

The equation is already in a standard form ${(\frac{x - x_{c}}{a})}^{2} + {(\frac{y - y_{c}}{b})}^{2}$ $((x-x_c)/a)^2+((y-y_c)/b)^2$ so the center $(x_{c}, y_{c})$ $(x_c,y_c)$ and semi-axes $a$ $a$ and $b$ $b$ can be read straight off it as $(3, 8)$ $(3,8)$ , $\frac{5}{3}$ $5/3$ and $1$ $1$ .

$a > b$ $a>b$ so the line through the foci (the major axis) is parallel to the $x$ $x$ -axis rather than the $y$ $y$ .

The vertices are at the center $\pm$ $+-$ semiaxes: $(3 \pm \frac{5}{3}, 8 \pm 1)$ $(3+-5/3,8+-1)$
The distance between each focus and the centre is $\sqrt{a^{2} - b^{2}}$ $sqrt(a^2-b^2$ = $\sqrt{(\frac{25}{9}) - 1^{2}} = \frac{4}{3}$ $sqrt((25/9)-1^2)=4/3$ . So the foci are $(3 \pm \frac{4}{3}, 8)$ $(3+-4/3,8)$

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How do you find the center, foci and vertices of $\frac{{(x - 3)}^{2}}{\frac{25}{9}} + {(y - 8)}^{2} = 1$ $(x-3)^2/(25/9)+(y-8)^2=1$ ?

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Explanation:

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How do you find the center, foci and vertices of (x−3)2259+(y−8)2=1(x-3)^2/(25/9)+(y-8)^2=1?

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How do you find the center, foci and vertices of $\frac{{(x - 3)}^{2}}{\frac{25}{9}} + {(y - 8)}^{2} = 1$ $(x-3)^2/(25/9)+(y-8)^2=1$ ?