Question

Find the standard form of the equation of the ellipse with the given characteristics. Vertices: (4, 0), (4, 14); endpoints of the minor axis: (1, 7), (7, 7)

Answer 1

Answer:

The standard form of the ellipse is $\frac{(x-4)^{2}}{9} + \frac{(y-7)^{2}}{49} = 1$.

Step-by-step explanation:

The major axis of the ellipse is located in the y axis, whereas the minor axis is in the x axis. The center of the ellipse is the midpoint of the line segment between vertices, this is:

$(h, k) =\frac{1}{2}\cdot V_{1} (x,y) + \frac{1}{2}\cdot V_{2} (x,y)$ (1)

If we know that $V_{1} (x,y) = (4,0)$ and $V_{2}(x,y) = (4, 14)$, then the coordinates of the center are, respectively:

$(h,k) = \frac{1}{2}\cdot (4, 0) + \frac{1}{2}\cdot (4,14)$

$(h,k) = (2,0) + (2, 7)$

$(h, k) = (4, 7)$

The length of each semiaxis is, respectively:

$a = \sqrt{(1 - 4)^{2}+(7-7)^{2}}$

$a = 3$

$b = \sqrt{(4-4)^{2}+(0-7)^{2}}$

$b = 7$

The standard equation of the ellipse is described by the following formula:

$\frac{(x-h)^{2}}{a^{2}}+ \frac{(y-k)^{2}}{b^{2}} = 1$

Where:

$h$, $k$ - Coordinates of the center of the ellipse.

$a$, $b$ - Length of the orthogonal semiaxes.

If we know that $h = 4$, $k = 7$, $a = 3$ and $b = 7$, then the standard form of the ellipse is:

$\frac{(x-4)^{2}}{9} + \frac{(y-7)^{2}}{49} = 1$