Question

Complete the following: by the di tant to Complete the squares for each quadratic, list the center and radius, then graph each circle (a-ha r-rs-i Since the radius is an imaginary value, the equation is not a real circle. (h) x2 + y2 - 7y = 0 rº + y2 + 2mr - 2ny = 0 letter h from question 1 please

Answer 1

EXPLANATION

Given the equation x^2 + y^2 - 7y = 0

As we already know, the Ellipse Standard Equation is as follows:

$\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$

With center (h,k) and (a,b) are the semi-major and semi-minor axes.

Rewrite x^2+y^2 -7y =0 in the form of the standard ellipse equation:

Group x-variables and y-variables together:

$x^2+(y^2-7y)=0$

Convert y to square form:

$x^2+(y^2-7y+\frac{49}{4})=0+\frac{49}{4}$

Refine 0+49/4

$x^2+(y-\frac{7}{2})^2=0+\frac{49}{4}$

Refine 0+49/4

$x^2+(y-\frac{7}{2})^2=\frac{49}{4}$

Divide by 49/4:

$\frac{x^2}{\frac{49}{4}}+\frac{(y-\frac{7}{2})^2}{\frac{49}{4}}=1$

Rewrite in standard form:

$\frac{(x-0)^2}{(\frac{7}{2})^2}+\frac{(y-\frac{7}{2})^2}{(\frac{7}{2})^2}=1$

Therefore, ellipse properties are:

(h,k)=(0,7/2) a=7/2, b=7/2

b>a therefore b is semi-major axis b=7/2, semi-minor axis a=7/2

The properties of the ellipse are: center (h,k)=(0,7/2), semi-major axis b=7/2 and semi-minor axis a=7/2.

Then, the graph is: