finding the standard form the center eccentricity vertices foci and minor axis endpoints of the equation 12x^2+4y^2-24x-4y+1=0

Please use proper English. I cannot understand what you are asking if you don't use a proper sentence.

I will complete the square.

12(x^2 - 24x +144) + 4(y^2 - y + 1/4) = -1+12(144)+4(1/4)
12(x-12)^2 + 4(y - 1/2)^2 = 1728
divide by 1728
(x-12)^2/144 + (y-1/2)^2/432 = 1

from there we know a = 12 , b = 12√3

find c, and take it from there

I factored incorrectly, let me try this again

12(x^2 - 2x + 1) + 4(y^2 - y + 1/4) = -1 + 12 - 1
12(x-1)^2 + 4(Y-1/2)^2 = 12
(x-1)^2 /1 + (y-1/2)^2 /3 = 1

etc

To find the standard form, center, eccentricity, vertices, foci, and minor axis endpoints of the given equation 12x^2 + 4y^2 - 24x - 4y + 1 = 0, we can follow these steps:

Step 1: Rearrange the equation
First, let's rearrange the equation by grouping the x-terms together and the y-terms together, completing the square for both x and y, and isolating the constant term on one side:

12x^2 - 24x + 4y^2 - 4y = -1

Step 2: Divide by the constant term
Next, we divide the entire equation by the common factor of the coefficients of x^2 and y^2 to make them 1:

x^2 - 2x + (4/12)y^2 - (4/4)y = -1/12

Simplifying further by reducing the fractions:

x^2 - 2x + (1/3)y^2 - y = -1/12

Step 3: Complete the square for x
Now, let's complete the square for x by adding the square of half the coefficient of x to both sides of the equation. In this case, the coefficient of x is -2, so we add (-2/2)^2 = 1 to both sides:

x^2 - 2x + 1 + (1/3)y^2 - y = -1/12 + 1

Simplifying:

(x - 1)^2 + (1/3)y^2 - y = 11/12

Step 4: Complete the square for y
Next, complete the square for y by adding the square of half the coefficient of y to both sides of the equation. The coefficient of y is -1, so we add (-1/2)^2 = 1/4 to both sides:

(x - 1)^2 + (1/3)y^2 - y + 1/4 = 11/12 + 1/4

Simplifying further:

(x - 1)^2 + (1/3)(y^2 - 3y + 3/4) = 11/12 + 3/12

(x - 1)^2 + (1/3)(y - 3/2)^2 = 14/12

Simplifying the right side:

(x - 1)^2 + (1/3)(y - 3/2)^2 = 7/6

Step 5: Convert to standard form
To convert the equation to standard form, we need to divide both sides of the equation by the constant on the right side (7/6) such that the equation equals 1:

[(x - 1)^2] / [(7/6)] + [(y - 3/2)^2] / [(7/18)] = 1

Simplifying further:

[(x - 1)^2] / [(7/6)] + [(y - 3/2)^2] / [(7/18)] = 1

This is the standard form of the equation.

Step 6: Identify the center, eccentricity, vertices, foci, and minor axis endpoints

- Center: The center of the ellipse can be found by looking at the values inside the parentheses. The center is at (h, k). So, the center of the given ellipse is (1, 3/2).

- Eccentricity: The eccentricity is the square root of the ratio of the major axis length to the minor axis length. In this case, the major axis length is 2 times the square root of (7/6), and the minor axis length is 2 times the square root of (7/18). By calculating this ratio, you can determine the eccentricity.

- Vertices: The vertices can be found by adding or subtracting the semi-major axis length from the x-coordinate of the center point. The semi-major axis length is the square root of [(7/6)]. So the vertices are (1 ± √(7/6), 3/2).

- Foci: The foci can be found using the formula c^2 = a^2 - b^2, where c represents the distance from the center to each focus point. a is the semi-major axis length, and b is the semi-minor axis length. By substituting the given values, you can calculate the foci points.

- Minor Axis Endpoints: Similarly, the minor axis endpoints can be found by adding or subtracting the semi-minor axis length from the y-coordinate of the center point. The semi-minor axis length is the square root of [(7/18)]. So the minor axis endpoints are (1, 3/2 ± √(7/18)).

By following these steps, you can find the standard form, center, eccentricity, vertices, foci, and minor axis endpoints of the given equation.