1. Find the distance of the line segment joining the two points:
(1/3,-2/3) and (-5/2,-9/4)
2. Identify the conic section represented by 6x^2+4y^2+24x-32y+64=0. Explain why.
3. The equation of a circle is: 6x^2-12x+6y^2+36y=36. Write the equation in standard form.
Wrong for all 3, what else is new ?
1) d^2 = (-2/3 + 9/4)^2 + (1/3 + 5/2)^2
= (19/12)^2 + (17/6)^2
= 361/144 + 289/36
= 1517/144
d = √1517 / 12 = appr 3.246
2. True, but the important part is for the ellipse A ≠ B, which was not stated
3. 6x^2-12x+6y^2+36y=36 , need completing the square routine
first divide all terms by 6
x^2 - 2x + y^2 + 6y = 6
x^2 - 2x + 1 + y^2 + 6y + 9 = 6 + 1 + 9
(x - 1)^2 + (y + 3)^2 = 16
so the circle has centre (1, -3) and radius of 4
Answer 1: The distance of the line segment joining the two points is sqrt(41/12).
Answer 2: The conic section represented by 6x^2+4y^2+24x-32y+64=0 is an ellipse. This is because the equation is in the form of Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0, where A, B, C, D, E, and F are all constants, and A and C are both positive. This is the standard form for an ellipse.
Answer 3: The equation of the circle in standard form is (x-6)^2+(y-18)^2=36.