Given an ellipse; 3x^2+5y^2-12x-50y+62=0, find the Center, the Major Axis, the Minor Axis, & the distance from C to Foci?
To find the center, major axis, minor axis, and the distance from the center to the foci of an ellipse, we first need to put the equation of the ellipse in standard form. The standard form is expressed as:
((x - h)^2 / a^2) + ((y - k)^2 / b^2) = 1
where (h, k) represents the center of the ellipse, and 'a' and 'b' represent the semi-major and semi-minor axes, respectively.
Let's go step-by-step to find the information you're looking for:
Step 1: Rearrange the given equation to bring it into standard form.
3x^2 + 5y^2 - 12x - 50y + 62 = 0
Rearranging the equation, we get:
3x^2 - 12x + 5y^2 - 50y = -62
Step 2: Complete the square for both x and y terms.
To complete the square for x terms, we need to add and subtract the square of half the coefficient of x:
3(x^2 - 4x + 4) + 5y^2 - 50y = -62 + 12
Applying the same process for y terms, we have:
3(x^2 - 4x + 4) + 5(y^2 - 10y + 25) = -62 + 12 + 5(25)
Simplifying the equation further:
3(x - 2)^2 + 5(y - 5)^2 = 25
Step 3: Divide both sides by 25 to make the right side equal to 1 (standard form).
[(x - 2)^2 / (5 / 5)] + [(y - 5)^2 / (25 / 5)] = 1
In the simplified form, we have:
(x - 2)^2 / 5 + (y - 5)^2 / 25 = 1
Step 4: Compare the equation to the standard form equation:
((x - h)^2 / a^2) + ((y - k)^2 / b^2) = 1
From the comparison, we can determine the values:
h = 2 (x-coordinate of the center)
k = 5 (y-coordinate of the center)
a^2 = 5 (square of the semi-major axis)
b^2 = 25 (square of the semi-minor axis)
Therefore, the center of the ellipse is (2, 5), the semi-major axis is √5 (approximately 2.236), the semi-minor axis is 5, and the distance from the center to the foci can be calculated using the formula:
c = √(a^2 - b^2)
c = √(5 - 25)
c = √(-20)
Since the square root of a negative number is not a real number, it implies that the given ellipse does not have any foci.