Question

1 fill in the blank 5 points identify the center, vertices, and co - vertices of the ellipse represented by $4x^{2}-16x + y^{2}-6y + 21 = 0$. list the vertices and co - vertices in order from left to right or bottom to top. center: ( type your answer..., type your answer...) vertices: ( type your answer..., type your answer...) and ( type your answer..., type your answer... co - vertices: ( type your answer..., type your answer...) and ( type your answer..., type your answe

Explanation:

Step1: Complete the square for x and y terms

First, group the x - terms and y - terms:

$$
LATEXBLOCK0
$$

For the \(x\) - terms: \(x^{2}-4x=(x - 2)^{2}-4\), so \(4(x^{2}-4x)=4((x - 2)^{2}-4)=4(x - 2)^{2}-16\)
For the \(y\) - terms: \(y^{2}-6y=(y - 3)^{2}-9\)

Substitute back into the equation:

$$
LATEXBLOCK1
$$

Divide both sides by 4 to get the standard form of the ellipse:

$$ \frac{(x - 2)^{2}}{1}+\frac{(y - 3)^{2}}{4}=1 $$

Step2: Identify the center, vertices, and co - vertices

The standard form of an ellipse is \(\frac{(x - h)^{2}}{b^{2}}+\frac{(y - k)^{2}}{a^{2}} = 1\) (when \(a>b\) and the major axis is vertical), where \((h,k)\) is the center, vertices are \((h,k\pm a)\), and co - vertices are \((h\pm b,k)\)

From \(\frac{(x - 2)^{2}}{1}+\frac{(y - 3)^{2}}{4}=1\), we have \(h = 2\), \(k = 3\), \(a=\sqrt{4}=2\), \(b=\sqrt{1}=1\)

• Center: \((h,k)=(2,3)\)
• Vertices: Since the major axis is vertical (because \(a^{2}=4\) is under the \(y\) - term), the vertices are \((2,3 + 2)=(2,5)\) and \((2,3-2)=(2,1)\)
• Co - vertices: Since the minor axis is horizontal (because \(b^{2}=1\) is under the \(x\) - term), the co - vertices are \((2 + 1,3)=(3,3)\) and \((2-1,3)=(1,3)\)

Answer:

Center: \((2,3)\)
Vertices: \((2,1)\) and \((2,5)\)
Co - vertices: \((1,3)\) and \((3,3)\)