Conics

Hello could anyone please tell me if the following answers to the problems are correct? if they are not could you please show me where I went wrong? Thanks!

1) Find the center, foci, vertices, and eccentricity of the of the ellipse and sketch graph of the follwing equation:

16x^(2) + 25y^(2) - 64x + 150y + 279= 0

Solution:
Center- (2, -3)
Vertices- ( 2 + 0.791, -3) ; (2 -0.791, -3)
Foci- ( 2 + 0.474, -3) ; (2 -0.474, -3)
eccentricity- 0.474
0.791

2) Find the equation of each ellipse given the following:

a) Vertices- (0, 2) ; (4, 2)
Eccentricity- 1
2

Solution: (x - 2)^(2) + (y - 2)^(2) = 1
4 3

b) Foci- (0, 5) ; (0, -5)
Major axis length- 14

Solution: x^(2) + y^(2) = 1
24 49

c) Center- (1, 2)
major axis- vertical
points on the ellipse: (1, 6) ; (3, 2)

Solution: (x -1)^(2) + (y -2)^(2) = 1
4 16

Just poking around the first one.

1) Why not put the answer in exact form? $\displaystyle \frac{1}{4}\sqrt{10}$ or $\displaystyle \frac{3}{20}\sqrt{10}$ or $\displaystyle \frac{1}{5}\sqrt{10}$ or $\displaystyle e\;= \;\frac{3}{5}$. Just look what it does for the expression of the eccentricity!

2) Aren't you missing two important points? Maybe the ends of the minor axis?

3) It looks like you have the right idea. Good work. Why are you asking for confirmation? Do you doubt?

Hello, googoogaga!

Quote:

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1) Find the center, foci, vertices, and eccentricity of the of the ellipse
and sketch graph of the follwing equation: .$\displaystyle 16x^2 + 25y^2 - 64x + 150y + 279 \:=\:0$

Solution
$\displaystyle \begin{array}{cc}\text{Center:} & (2,\,-3) \\
\text{Vertices:} & (2 \pm0.791,\,-3) \\
\text{Foci:} & (2 \pm 0.474,\,-3) \\
\text{Eccentricity: } & \frac{0.474}{0.791}\end{array}$



2) Find the equation of each ellipse given the following:

$\displaystyle a)\;\begin{array}{cc}\text{Vertices:} & (0,\, 2),\;(4,\,2)\\
\text{Eccentricity:} & \frac{1}{2}\end{array}$

Solution:. . $\displaystyle \frac{(x - 2)^2}{4} + \frac{(y - 2)^2}{3} \;=\;1$



$\displaystyle b)\;\begin{array}{cc}\text{Foci:} & (0,\,\pm5)\\
\text{Major axis:} & 14\end{array}$

Solution: .$\displaystyle \frac{x^2}{24} + \frac{y^2}{49} \;=\;1$



$\displaystyle c)\;\begin{array}{cc}\text{Center:} & (1,\,2) \\
\text{Major axis:} & \text{vertical} \\
\text{points on ellipse:} & (1,\,6),\;(3,\,2) \end{array}$

Solution: .$\displaystyle \frac{(x -1)^2}{4} + \frac{(y -2)^2}{16} \;=\;1$

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Looks good to me . . . Good work!