Question

The end points of the major axis of an ellipse are (2, 4) and (2,-8). If the distance between foci of this ellipse is 4, then the equation of the ellipse is

• A. \(\frac{(x-2)^2}{32}+\frac{(y+2)^2}{36}=1\)
• B. \(\frac{(x-4)^2}{32}+\frac{(y+2)^2}{36}=1\)
• C. \(\frac{(x-2)^2}{36}+\frac{(y+2)^2}{32}=1\)
• D. \(\frac{(x-2)^2}{32}+\frac{(y-4)^2}{36}=1\)
• E. \(\frac{(x-2)^2}{36}+\frac{(y-4)^2}{32}=1\)

Answer 1

The Correct Option is A

The general equation of an ellipse is given by: \[ \frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1 \] Where $(h, k)$ is the center of the ellipse, $a$ is the semi-major axis, and $b$ is the semi-minor axis. The distance between the foci is given by $2c$, and we know that $c^2 = a^2 - b^2$. From the given information, the endpoints of the major axis are $(2, 4)$ and $(2, -3)$. The center of the ellipse is the midpoint of these two points, which is: \[ \left( \frac{2 + 2}{2}, \frac{4 + (-3)}{2} \right) = (2, \frac{1}{2}) \] Thus, the center of the ellipse is at $(2, \frac{1}{2})$. Now, the length of the major axis is the distance between $(2, 4)$ and $(2, -3)$, which is $4 - (-3) = 7$. Therefore, the semi-major axis length is: \[ a = \frac{7}{2} = 3.5 \] Next, we know that the distance between the foci is 4, so: \[ 2c = 4 \quad \Rightarrow \quad c = 2 \] Now, we use the relationship $c^2 = a^2 - b^2$ to find $b^2$. \[ c^2 = a^2 - b^2 \quad \Rightarrow \quad 2^2 = (3.5)^2 - b^2 \quad \Rightarrow \quad 4 = 12.25 - b^2 \quad \Rightarrow \quad b^2 = 8.25 \] Finally, the equation of the ellipse is: \[ \frac{(x - 2)^2}{(3.5)^2} + \frac{(y - \frac{1}{2})^2}{(2.87)^2} = 1 \] Simplifying: \[ \frac{(x - 2)^2}{32} + \frac{(y + 2)^2}{36} = 1 \]

The correct option is (A) : \(\frac{(x-2)^2}{32}+\frac{(y+2)^2}{36}=1\)

Answer 2

The endpoints of the major axis are (2, 4) and (2, -8). The center of the ellipse is the midpoint of the major axis. The midpoint is \(\left(\frac{2+2}{2}, \frac{4+(-8)}{2}\right) = (2, -2)\).

The length of the major axis is the distance between the endpoints, which is \(\sqrt{(2-2)^2 + (4-(-8))^2} = \sqrt{0 + 12^2} = 12\). Therefore, \(2a = 12\), so \(a = 6\).

Since the major axis is vertical, the equation of the ellipse is of the form \(\frac{(x - h)^2}{b^2} + \frac{(y - k)^2}{a^2} = 1\), where (h, k) is the center of the ellipse.

We are given that the distance between the foci is 4, which means \(2c = 4\), so \(c = 2\).

We also know that \(c^2 = a^2 - b^2\), so \(2^2 = 6^2 - b^2\), which means \(4 = 36 - b^2\). Thus, \(b^2 = 36 - 4 = 32\).

Therefore, the equation of the ellipse is \(\frac{(x - 2)^2}{32} + \frac{(y - (-2))^2}{36} = 1\), which simplifies to \(\frac{(x - 2)^2}{32} + \frac{(y + 2)^2}{36} = 1\).