Question

determine the eccentricity of the ellipse given by $9x^{2}+16y^{2}-72x + 64y-368 = 0$.
0.88
0.60
1.51
0.66
question 8 (5 points)
write the standard form of the equation for the circle that passes through the points $(2,31),(-15,14),(33,0)$. then identify the center and radius.
$(x - 6)^{2}+(y - 5)^{2}=625$; center $(6,5),r = 25$
$(x - 9)^{2}+(y - 7)^{2}=625$; center $(9,7),r = 25$
$(x + 7)^{2}+(y + 10)^{2}=25$; center $(7,10),r = 25$
$(x + 7)^{2}+(y + 10)^{2}=25$; center $(7,10),r = 5$

Explanation:

Step1: Complete the square for the ellipse equation

Given \(9x^{2}+16y^{2}-72x + 64y-368 = 0\).
Group \(x\) - terms and \(y\) - terms:
\(9(x^{2}-8x)+16(y^{2}+4y)=368\).
Complete the square for \(x\): \(x^{2}-8x=(x - 4)^{2}-16\).
Complete the square for \(y\): \(y^{2}+4y=(y + 2)^{2}-4\).
Substitute back:
\(9((x - 4)^{2}-16)+16((y + 2)^{2}-4)=368\).
Expand: \(9(x - 4)^{2}-144+16(y + 2)^{2}-64 = 368\).
Simplify: \(9(x - 4)^{2}+16(y + 2)^{2}=368 + 144+64=576\).
Divide by \(576\): \(\frac{(x - 4)^{2}}{64}+\frac{(y + 2)^{2}}{36}=1\).
Here \(a^{2}=64\), so \(a = 8\); \(b^{2}=36\), so \(b = 6\).
Use the formula \(c=\sqrt{a^{2}-b^{2}}\), then \(c=\sqrt{64 - 36}=\sqrt{28}=2\sqrt{7}\).
The eccentricity \(e=\frac{c}{a}\), so \(e=\frac{2\sqrt{7}}{8}=\frac{\sqrt{7}}{4}\approx0.66\).

Step2: Check the circle equation

For a circle \((x - h)^{2}+(y - k)^{2}=r^{2}\).
If \((x - 6)^{2}+(y - 5)^{2}=625\) (center \((6,5)\), \(r = 25\)):
For point \((2,31)\): \((2 - 6)^{2}+(31 - 5)^{2}=16 + 676=692
eq625\).
If \((x - 9)^{2}+(y - 7)^{2}=625\) (center \((9,7)\), \(r = 25\)):
For point \((2,31)\): \((2 - 9)^{2}+(31 - 7)^{2}=49+576 = 625\).
For point \((-15,14)\): \((-15 - 9)^{2}+(14 - 7)^{2}=(-24)^{2}+7^{2}=576 + 49=625\).
For point \((33,0)\): \((33 - 9)^{2}+(0 - 7)^{2}=24^{2}+49=576+49 = 625\).

Answer:

For the ellipse, the eccentricity is \(0.66\).
For the circle, the equation is \((x - 9)^{2}+(y - 7)^{2}=625\), center \((9,7)\), \(r = 25\).