" 24.यद्वि "(4x^(2)-3y^(2)):(2x^(2)+5y^(2))=12:19" हो,तो "x:y=?

If (4x^2-3y^2):(2x^2+5y^2)=12:19 then x:y=

Solve for (x - 1)^(2) and (y + 3)^(2) , 2x^(2) - 5y^(2) - x - 27y - 26 = 3(x + y + 5) and 4x^(2) - 3y^(2) - 2xy + 2x - 32y - 16 = (x - y + 4)^(2) .

Identify the type of conic section for each of the equations 1. 2x^(2) -y^(2) = 7 2. 3x^(2) +3 y^(2) -4x + 3y + 10 =0 3. 3x^(2) + 2y^(2) = 14 4. x^(2) + y^(2) + x-y=0 5. 11x^(2) -25y^(2) -44x + 50y - 256 =0 6. y^(2) + 4x + 3y + 4=0

Find each of the following products: (i) (4x + 5y) (4x - 5y) (ii) (3x^(2) + 2y^(2)) (3x^(2) - 2y^(2))

Find the radical center of the circles x^(2)+y^(2)+4x+6y=19,x^(2)+y^(2)=9,x^(2)+y^(2)-2x-4y=5

3x^(2) + 8x +4=0, 4y^(2)-19y + 12=0

Solve for x and y:3x^(2)-5y^(2)=7,3xy-4y^(2)=2