Question

1. write the equation in standard form. then identify the conic section and its center.

$x^{2}+y^{2}-8x + 10y+15 = 0$
part i: complete the square to write this equation in standard form. (4 points)
part ii: what conic section does this equation define? (2 points)
part iii: what is the center, or vertex, of this conic section? (2 points)

Explanation:

Step1: Rearrange and complete the square for x - terms

Group the x - terms and y - terms: $(x^{2}-8x)+(y^{2}+10y)+15 = 0$.
For the x - terms: $x^{2}-8x=(x - 4)^{2}-16$.
For the y - terms: $y^{2}+10y=(y + 5)^{2}-25$.
The equation becomes $(x - 4)^{2}-16+(y + 5)^{2}-25+15 = 0$.

Step2: Simplify the equation

Combine the constants: $(x - 4)^{2}+(y + 5)^{2}-26 = 0$.
Then, $(x - 4)^{2}+(y + 5)^{2}=26$.

Step3: Identify the conic - section

The general form of a circle is $(x - h)^{2}+(y - k)^{2}=r^{2}$, where $(h,k)$ is the center and $r$ is the radius. Since our equation is in the form $(x - 4)^{2}+(y + 5)^{2}=26$, it represents a circle.

Step4: Find the center

Comparing $(x - 4)^{2}+(y + 5)^{2}=26$ with $(x - h)^{2}+(y - k)^{2}=r^{2}$, we can see that $h = 4$ and $k=-5$. So the center of the circle is $(4,-5)$.

Answer:

Part I: $(x - 4)^{2}+(y + 5)^{2}=26$
Part II: Circle
Part III: $(4,-5)$