MATH SOLVE

4 months ago

Q:
# help me please...Problem: The standard form of a circle is (x-h)2+(y-k)2=r2 and for the parabola, y-k=a(x-h)2. The (h,k) pair will be the center of the circle and the vertex of the parabola. The radius of the circle is ‘r’ and the focal length of the parabola is f=1/(4a). For the following General Conic Equation: x2+y2-4x-6y-12=0 complete the following problems showing all your work: Complete the square showing all your work to convert to Standard Form: If this is a circle, state the coordinates of the center and give the radius. If this is a parabola, state the coordinates of the vertex and give the focal length. Show all your work. Sketch the Conic. Label the values you found in part B. Be sure to draw or show the radius or focal length.

Accepted Solution

A:

we have that

x²+y²-4x-6y-12=0

Group terms that contain the same variable, and move the constant to the opposite side of the equation

(x²-4x)+(y²-6y)=12

Complete the square twice. Remember to balance the equation by adding the same constants to each side

(x²-4x+4)+(y²-6y+9)=12+4+9

Rewrite as perfect squares

(x-2)²+(y-3)²=25

the answer part A) is

(x-2)²+(y-3)²=5²-----> this is the standard form of the equation of a circle

Part B) (x-2)²+(y-3)²=5²

the center is the point (2,3) and the radius is r=5 units

Part C)

using a graph tool

see the attached figure

x²+y²-4x-6y-12=0

Group terms that contain the same variable, and move the constant to the opposite side of the equation

(x²-4x)+(y²-6y)=12

Complete the square twice. Remember to balance the equation by adding the same constants to each side

(x²-4x+4)+(y²-6y+9)=12+4+9

Rewrite as perfect squares

(x-2)²+(y-3)²=25

the answer part A) is

(x-2)²+(y-3)²=5²-----> this is the standard form of the equation of a circle

Part B) (x-2)²+(y-3)²=5²

the center is the point (2,3) and the radius is r=5 units

Part C)

using a graph tool

see the attached figure