SOLUTION: 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

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.
RELATED QUESTIONS
what is the standard form equation of the conic given ? 4x^2-5y^2-16x-30y-9=... (answered by stanbon)

What is the standard form of the conic with the equation:... (answered by stanbon)

How do I go about converting this equation to the standard form in order to graph the... (answered by rapaljer)

Identify the conic(cirlce, ellipse, hyperbola, ellipse. Write each of the following in... (answered by jim_thompson5910)

Write the equation for 5x^2+5y^2-40x+30y-120=0 in standard form. Show work. (answered by josgarithmetic)

put the equation 16x^2+y^2+64x-2y+67=0 into conic standard... (answered by Edwin McCravy,MathLover1,ikleyn)

What is the standard form of the equation of the conic:5x^2-4y^2-30x-16y+9=0 (answered by Fombitz)

what is the latus rectum of this equation? y^2-12y+16x+36=0 standard form =... (answered by lwsshak3)

What is the standard form of... (answered by MathLover1)