Question

1. determine the center, vertices, co - vertices, and the foci for the following $\frac{x^{2}}{25}+\frac{y^{2}}{16}=1$. 2. determine the center, vertices, co - vertices, and the foci for the following $\frac{(x - 2)^{2}}{9}+\frac{(y + 1)^{2}}{36}=1$. 3. determine the center, vertices, co - vertices, and the foci for the following $\frac{(x + 3)^{2}}{14}+\frac{y^{2}}{49}=1$.

Explanation:

Step1: Recall standard - form of ellipse equation

The standard form of an ellipse centered at $(h,k)$ is $\frac{(x - h)^2}{a^2}+\frac{(y - k)^2}{b^2}=1$ (for a horizontally - oriented ellipse) or $\frac{(x - h)^2}{b^2}+\frac{(y - k)^2}{a^2}=1$ (for a vertically - oriented ellipse), where $a>b>0$, the center is $(h,k)$, vertices are $(h\pm a,k)$ or $(h,k\pm a)$ depending on orientation, co - vertices are $(h\pm b,k)$ or $(h,k\pm b)$ depending on orientation, and $c=\sqrt{a^2 - b^2}$ and foci are $(h\pm c,k)$ or $(h,k\pm c)$ depending on orientation.

Problem 1: $\frac{x^2}{25}+\frac{y^2}{16}=1$

Step1: Identify the center

Since the equation is of the form $\frac{(x - 0)^2}{25}+\frac{(y - 0)^2}{16}=1$, the center $(h,k)=(0,0)$.

Step2: Identify $a$ and $b$

Here $a^2 = 25$, so $a = 5$ and $b^2=16$, so $b = 4$.

Step3: Find the vertices

Since $a^2$ is under $x^2$, the ellipse is horizontally oriented. The vertices are $(h\pm a,k)=( \pm 5,0)$.

Step4: Find the co - vertices

The co - vertices are $(h\pm b,k)=(0,\pm 4)$.

Step5: Find $c$ and the foci

$c=\sqrt{a^2 - b^2}=\sqrt{25 - 16}=\sqrt{9}=3$. The foci are $(h\pm c,k)=( \pm 3,0)$.

Problem 2: $\frac{(x - 2)^2}{9}+\frac{(y + 1)^2}{36}=1$

Step1: Identify the center

The center $(h,k)=(2,-1)$.

Step2: Identify $a$ and $b$

Here $a^2 = 36$, so $a = 6$ and $b^2 = 9$, so $b = 3$.

Step3: Find the vertices

Since $a^2$ is under $(y + 1)^2$, the ellipse is vertically oriented. The vertices are $(h,k\pm a)=(2,-1\pm6)=(2,5)$ and $(2,-7)$.

Step4: Find the co - vertices

The co - vertices are $(h\pm b,k)=(2\pm3,-1)$, i.e., $(5,-1)$ and $(-1,-1)$.

Step5: Find $c$ and the foci

$c=\sqrt{a^2 - b^2}=\sqrt{36 - 9}=\sqrt{27}=3\sqrt{3}$. The foci are $(h,k\pm c)=(2,-1\pm3\sqrt{3})$.

Problem 3: $\frac{(x + 3)^2}{14}+\frac{y^2}{49}=1$

Step1: Identify the center

The center $(h,k)=(-3,0)$.

Step2: Identify $a$ and $b$

Here $a^2 = 49$, so $a = 7$ and $b^2 = 14$, so $b=\sqrt{14}$.

Step3: Find the vertices

Since $a^2$ is under $y^2$, the ellipse is vertically oriented. The vertices are $(h,k\pm a)=(-3,\pm7)$.

Step4: Find the co - vertices

The co - vertices are $(h\pm b,k)=(-3\pm\sqrt{14},0)$.

Step5: Find $c$ and the foci

$c=\sqrt{a^2 - b^2}=\sqrt{49 - 14}=\sqrt{35}$. The foci are $(h,k\pm c)=(-3,\pm\sqrt{35})$.

Answer:

Problem 1

Center: $(0,0)$
Vertices: $(5,0),(-5,0)$
Co - vertices: $(0,4),(0,-4)$
Foci: $(3,0),(-3,0)$

Problem 2

Center: $(2,-1)$
Vertices: $(2,5),(2,-7)$
Co - vertices: $(5,-1),(-1,-1)$
Foci: $(2,-1 + 3\sqrt{3}),(2,-1-3\sqrt{3})$

Problem 3

Center: $(-3,0)$
Vertices: $(-3,7),(-3,-7)$
Co - vertices: $(-3+\sqrt{14},0),(-3-\sqrt{14},0)$
Foci: $(-3,\sqrt{35}),(-3,-\sqrt{35})$