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Question 389516: It asks for the standard form of the equation of this conic:
4x^2 - 5y^2 - 16x - 30y - 9 = 0
I know the standard form is going to be the form of a hyperbola, since a and c are not equal and have opposite signs.
I also know that completing the square is part of solving this too, I just don't really know how to apply it to a conic.
Help is much appreciated!
Answer by haileytucki(390)   (Show Source):
You can put this solution on YOUR website! To find the properties of the Conic solution:
4x^(2)-5y^(2)-16x-30y-9=0
Move all terms not containing a variable to the right-hand side of the equation.
4x^(2)-16x-5y^(2)-30y=9
To create a trinomial square on the left-hand side of the equation, add a value to both sides of the equation that is equal to the square of half the coefficient of x. In this problem, add (-2)^(2) to both sides of the equation.
4(x^(2)-4x+4)-5y^(2)-30y=9+0+4*4
Factor the perfect trinomial square into (x-2)^(2).
4((x-2)^(2))-5y^(2)-30y=9+0+4*4
Factor the perfect trinomial square into (x-2)^(2).
4(x-2)^(2)-5y^(2)-30y=9+0+4*4
To create a trinomial square on the left-hand side of the equation, add a value to both sides of the equation that is equal to the square of half the coefficient of y. In this problem, add (3)^(2) to both sides of the equation.
4(x-2)^(2)-5(y^(2)+6y+9)=9+0+4*4+0+9*-5
Factor the perfect trinomial square into (y+3)^(2).
4(x-2)^(2)-5((y+3)^(2))=9+0+4*4+0+9*-5
Factor the perfect trinomial square into (y+3)^(2).
4(x-2)^(2)-5(y+3)^(2)=9+0+4*4+0+9*-5
Multiply 4 by 4 to get 16.
4(x-2)^(2)-5(y+3)^(2)=9+0+16+0+9*-5
Multiply 9 by -5 to get -45.
4(x-2)^(2)-5(y+3)^(2)=9+0+16+0-45
Add 0 to 9 to get 9.
4(x-2)^(2)-5(y+3)^(2)=9+16+0-45
Add 16 to 9 to get 25.
4(x-2)^(2)-5(y+3)^(2)=25+0-45
Add 0 to 25 to get 25.
4(x-2)^(2)-5(y+3)^(2)=25-45
Subtract 45 from 25 to get -20.
4(x-2)^(2)-5(y+3)^(2)=-20
Multiply each term in the equation by -1.
(4(x-2)^(2)-5(y+3)^(2))*-1=-20*-1
Multiply (4(x-2)^(2)-5(y+3)^(2)) by -1 to get -(4(x-2)^(2)-5(y+3)^(2)).
-(4(x-2)^(2)-5(y+3)^(2))=-20*-1
Multiply -20 by -1 to get 20.
-(4(x-2)^(2)-5(y+3)^(2))=20
Divide each term by 20 to make the right-hand side equal to 1.
-(4(x-2)^(2)-5(y+3)^(2))/(20)=(20)/(20)
Simplify each term in the equation in order to set the right-hand side equal to 1. The standard form of an ellipse or hyperbola requires the right-hand side of the equation be 1.
-(4(x-2)^(2)-5(y+3)^(2))/(20)=1
Let me know if this helps; if not, repost.
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