identify the type of conic section represented by 9x^2-25y^-18x+50y=0?

Answer 1

The best way to determine what type of conic section this is, is through algebraic manipulation. You have to complete the square with the x variables and the y variables.

9x^2 - 25y^2 - 18x + 50y = 0

Group the x and y variables together,

9x^2 - 18x - 25y^2 + 50y = 0

Now, you complete the square with each variable. I won't show you the detailed steps in how to complete the square, but that's essentially how to algebraically determine what conic section this is.

9(x^2 - 2x) - 25(y^2 + 2y) = 0

Adding a 1 in the first set of brackets, a 1 in the second set of brackets, and a 9 - 25 on the right hand side, we get

9(x^2 - 2x + 1) - 25(y^2 + 2y + 1) = 9 - 25

Simplifying this, as the bracketed terms are now binomial squares,

9(x - 1)^2 - 25(y + 1)^2 = -16

And then dividing everything by -16, we get

(-9/16)(x - 1)^2 + (25/16)(y + 1)^2 = 1

Since the first term is negative, it's best to swap the two expressions on the left hand side.

(25/16)(y + 1)^2 - (9/16)(x - 1)^2 = 1

And we can move the fraction to the bottom as division, to obtain

[(y + 1)^2]/(16/25) - [(x - 1)^2]/(16/9) = 1

The minus sign is a key indicator that this is a hyperbola. The fact that the y variable comes first indicates this to be a hyperbola that opens up and down. Also, the foci of the hyperbola is at (1, -1), and the asymptotes are determined by the width of the square roots of the denominators (4/3 lengthwise and 4/5 width).

Answer 2

It is a hyperbola. Complete the square for both x and y.

9x^2-18x+9-9-25y^2+50y-25+25=0
9(x^2-2x+1)-9-25(y^2-2y+1)+25=0
9(x-1)^2-25(y-1)^2+16=0
25(y-1)^2-9(x-1)^2=16

Answer 3

you may in the present day tell that's a circle because of the fact x² and y² have the same sign and denominators then x² + 8x + (4)² + y² - 4y + (-2)² = -11 + sixteen + 4 (x + 4)² + (y - 2)² = 9 so midsection: (-4, 2) radius: 3

Answer 4

9x^2-18x-25y^2+50y=0
9(x^2-2x+1)-25(y^2-2y+1)=-16
-(x-1)^2/(4/3)^2+(y-1)^2/(4/5)^2=1
so it is a hyperbola