Question

for questions 1 - 5, identify the conic section from its equation. (4 points each)

1. $\frac{(x - 5)^2}{4}+\frac{(y + 3)^2}{16}=1$
2. $\frac{(x - 4)^2}{36}-\frac{(y - 2)^2}{9}=1$
3. $x^{2}+y^{2}+6x - 16y + 48 = 0$
4. $y^{2}+6y - 2x + 13 = 0$
5. $\frac{(y - 2)^2}{4}-\frac{(x + 3)^2}{9}=1$

Explanation:

Step1: Recall conic - section equations

The standard form of an ellipse is $\frac{(x - h)^2}{a^2}+\frac{(y - k)^2}{b^2}=1$ ($a
eq b$), a hyperbola is $\frac{(x - h)^2}{a^2}-\frac{(y - k)^2}{b^2}=1$ or $\frac{(y - k)^2}{a^2}-\frac{(x - h)^2}{b^2}=1$, a circle is $(x - h)^2+(y - k)^2=r^2$ (which can be written as $x^{2}+y^{2}+Dx + Ey+F = 0$ with $D^{2}+E^{2}-4F>0$), and a parabola is either $y=ax^{2}+bx + c$ or $x = ay^{2}+by + c$.

Step2: Analyze equation 1

The equation $\frac{(x - 5)^2}{4}+\frac{(y + 3)^2}{16}=1$ is in the form of an ellipse $\frac{(x - h)^2}{a^2}+\frac{(y - k)^2}{b^2}=1$ where $h = 5,k=-3,a = 2,b = 4$.

Step3: Analyze equation 2

The equation $\frac{(x - 4)^2}{36}-\frac{(y - 2)^2}{9}=1$ is in the form of a hyperbola $\frac{(x - h)^2}{a^2}-\frac{(y - k)^2}{b^2}=1$ with $h = 4,k = 2,a = 6,b = 3$.

Step4: Analyze equation 3

Complete the square for $x^{2}+y^{2}+6x-16y + 48=0$.
$(x^{2}+6x)+(y^{2}-16y)=-48$.
$(x + 3)^2-9+(y - 8)^2-64=-48$.
$(x + 3)^2+(y - 8)^2=25$, which is a circle with center $(-3,8)$ and radius $r = 5$.

Step5: Analyze equation 4

Solve $y^{2}+6y-2x + 13=0$ for $x$.
$y^{2}+6y+9=2x - 4$.
$(y + 3)^2=2x - 4$, which is a parabola of the form $(y - k)^2=4p(x - h)$ where $k=-3,h = 2,p=\frac{1}{2}$.

Step6: Analyze equation 5

The equation $\frac{(y - 2)^2}{4}-\frac{(x + 3)^2}{9}=1$ is in the form of a hyperbola $\frac{(y - k)^2}{a^2}-\frac{(x - h)^2}{b^2}=1$ with $h=-3,k = 2,a = 2,b = 3$.

Answer:

1. Ellipse
2. Hyperbola
3. Circle
4. Parabola
5. Hyperbola