`y^2+5y-36` by `y+9`

y ^ (2) + 5y-36 was y + 9

The equations of the latus recta of the ellipse 9x^(2) + 25y^(2) - 36x + 50y - 164 = 0 are

Divide: 5/9\ b y2/3 (ii) 28\ b y7/4 36\ b y\ 6 2/3

Find the equation to the ausiliary circle of the ellipse 4x^(2) +9y^(2) -24x - 36 y + 36 = 0 .

Simplify : 5y(3y + 9)

(6,0),(0,6),a n d(7,7) are the vertices of a A B C . The incircle of the triangle has equation. (a) x^2+y^2-9x-9y+36=0 (b) x^2+y^2+9x-9y+36=0 (c) x^2+y^2+9x+9y-36=0 (d) x^2+y^2+18 x-18 y+36=0

(6,0),(0,6),a n d(7,7) are the vertices of a A B C . The incircle of the triangle has equation. a x^2+y^2-9x-9y+36=0 b x^2+y^2+9x-9y+36=0 c x^2+y^2+9x+9y-36=0 d x^2+y^2+18 x-18 y+36=0

Image of ellipse 4x^2 + 9y^2 = 36 in the line y=x is : (A) 9x^2 + 4y^2 = 36 (B) 3x^2 + 2y^2 = 36 (C) 2x^2+3y^2 = 36 (D) none of these