Question

Find the center, vertices, and foci of the ellipse with equation x squared divided by one hundred plus y squared divided by thirty six = 1

Answer 1

Answer on Question #45379 – Math – Analytical Geometry

Find the center, vertices, and foci of the ellipse with equation $x^2 / 100 + y^2 / 36 = 1$.

**Solution:**

$$\frac{x^2}{100} + \frac{y^2}{36} = 1$$

Since $x^2 = (x - 0)^2$ and $y^2 = (y - 0)^2$, the equation above is really:

$$\frac{(x - 0)^2}{100} + \frac{(y - 0)^2}{36} = 1$$

Then the center is at $(h, k) = (0, 0)$. I know that the $a^2$ is always the larger denominator (and $b^2$ is the smaller denominator), and this larger denominator is under the variable that parallels the longer direction of the ellipse. Since 100 is larger than 36, then $a^2 = 100$, $a = \pm \sqrt{100} = \pm 10$, and this ellipse is wider (paralleling the x-axis) than it is tall. The value of $a$ also tells me that the vertices are 10 units to either side of the center, at $(-10, 0)$ and $(10, 0)$.

Let's find co-vertices of the ellipse:

$$\begin{array}{l} b^2 = 36 \\ b = \pm \sqrt{3}6 = \pm 6 \end{array}$$

Co-vertices: $(-6, 0)$ and $(6, 0)$.

To find the foci, we need to find the value of $c$. From the equation, I already have $a^2$ and $b^2$, so:

$$\begin{array}{l} a^2 - c^2 = b^2 \\ 100 - c^2 = 36 \\ c^2 = 64 \\ c = \pm \sqrt{64} = \pm 8 \end{array}$$

Then the value of $c$ is 8, and the foci are eight units to either side of the center, at $(-8, 0)$ and $(8, 0)$.

**Answer:** center $(0,0)$, vertices are $(-10, 0)$ and $(10, 0)$, $(-6, 0)$ and $(6, 0)$. foci are $(-8,0)$ and $(8,0)$.

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