Question

1. $\begin{cases}x^{2}+y^{2}-16y + 39 = 0\\y^{2}-x^{2}-9 = 0end{cases}$ part i: what two shapes intersect in this problem? (2 points) part ii: solve the second equation for $x^{2}$. (4 points) part iii: use substitution to solve for two values of y. (4 points) part iv: use the values of y you just found to get the corresponding values of x, representing the points of intersection of these two shapes. write your answers in the form (x,y). (4 points)

Explanation:

Part I:

The first equation $x^{2}+y^{2}-16y + 39=0$ can be rewritten in the standard - form of a circle equation. Completing the square for the $y$ terms: $x^{2}+(y - 8)^{2}-64 + 39=0$, or $x^{2}+(y - 8)^{2}=25$, which is a circle with center $(0,8)$ and radius $r = 5$. The second equation $y^{2}-x^{2}-9=0$ can be rewritten as $y^{2}-x^{2}=9$, which is a hyperbola. So the two shapes are a circle and a hyperbola.

Part II:

Starting with the second equation $y^{2}-x^{2}-9 = 0$, solve for $x^{2}$:
$x^{2}=y^{2}-9$

Part III:

Substitute $x^{2}=y^{2}-9$ into the first equation $x^{2}+y^{2}-16y + 39=0$:
$(y^{2}-9)+y^{2}-16y + 39=0$
$2y^{2}-16y+30 = 0$
Divide through by 2: $y^{2}-8y + 15=0$
Factor the quadratic equation: $(y - 3)(y - 5)=0$
Set each factor equal to zero:
If $y-3=0$, then $y = 3$; if $y - 5=0$, then $y = 5$

Part IV:

When $y = 3$:
Substitute $y = 3$ into $x^{2}=y^{2}-9$, we get $x^{2}=3^{2}-9=0$, so $x = 0$. The point is $(0,3)$.
When $y = 5$:
Substitute $y = 5$ into $x^{2}=y^{2}-9$, we get $x^{2}=5^{2}-9=16$, so $x=\pm4$. The points are $(4,5)$ and $(-4,5)$

Answer:

Part I: A circle and a hyperbola
Part II: $x^{2}=y^{2}-9$
Part III: $y = 3,y = 5$
Part IV: $(0,3),(4,5),(-4,5)$