Question

question 9 of 10
which conic section does this equation represent?
$x^{2}-4x - y^{2}+6y - 9 = 5$
a. ellipse
b. circle
c. parabola
d. hyperbola

Explanation:

Step1: Complete the square for x and y terms

First, group the x - terms and y - terms:
$$(x^{2}-4x)-(y^{2}-6y)=5 + 9$$
For the x - terms: $x^{2}-4x=(x - 2)^{2}-4$. For the y - terms: $y^{2}-6y=(y - 3)^{2}-9$.
So, $(x - 2)^{2}-4-((y - 3)^{2}-9)=14$.

Step2: Simplify the equation

Expand the left - hand side:
$$(x - 2)^{2}-4-(y - 3)^{2}+9 = 14$$
$$(x - 2)^{2}-(y - 3)^{2}=14 - 9+4$$
$$(x - 2)^{2}-(y - 3)^{2}=9$$
The general form of a hyperbola is $\frac{(x - h)^{2}}{a^{2}}-\frac{(y - k)^{2}}{b^{2}} = 1$ (for a hyperbola opening left - right) or $\frac{(y - k)^{2}}{a^{2}}-\frac{(x - h)^{2}}{b^{2}} = 1$ (for a hyperbola opening up - down). Our equation $(x - 2)^{2}-(y - 3)^{2}=9$ can be written as $\frac{(x - 2)^{2}}{9}-\frac{(y - 3)^{2}}{9}=1$, which is in the form of a hyperbola.

Answer:

D. Hyperbola