Question

1. $\frac{(x + 2)^2}{9}+\frac{(y - 3)^2}{16}=1$ part i: what kind of conic section is this? (2 points) part ii: what are the coordinates of its center? (2 points) part iii: identify the coordinates of the four vertices of this graph. (2 points) part iv: identify the coordinates of the foci of this graph. (2 points)

Explanation:

Step1: Identify conic - section type

The general form of an ellipse is $\frac{(x - h)^2}{a^2}+\frac{(y - k)^2}{b^2}=1$. Given $\frac{(x + 2)^2}{9}+\frac{(y - 3)^2}{16}=1$, it is an ellipse.

Step2: Find center coordinates

For an ellipse in the form $\frac{(x - h)^2}{a^2}+\frac{(y - k)^2}{b^2}=1$, the center is $(h,k)$. Here, $h=-2$ and $k = 3$, so the center is $(-2,3)$.

Step3: Find vertices

Since $b^2=16>9 = a^2$, the major - axis is vertical. The vertices are $(h,k\pm b)$. Here, $a = 3$, $b = 4$, so the vertices are $(-2,3 + 4)=(-2,7)$ and $(-2,3 - 4)=(-2,-1)$. The other two vertices (ends of the minor - axis) are $(h\pm a,k)$, i.e., $(-2+3,3)=(1,3)$ and $(-2 - 3,3)=(-5,3)$.

Step4: Find foci

The relationship for an ellipse is $c^2=b^2 - a^2$. Here, $c^2=16 - 9=7$, so $c=\sqrt{7}$. The foci are $(h,k\pm c)$, i.e., $(-2,3+\sqrt{7})$ and $(-2,3-\sqrt{7})$.

Answer:

Part I: Ellipse
Part II: $(-2,3)$
Part III: $(-2,7),(-2,-1),(1,3),(-5,3)$
Part IV: $(-2,3+\sqrt{7}),(-2,3-\sqrt{7})$