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What is the standard form of the equation below?
 

5x (to the second power) + 3y (to the second power) - 20x + 24y = -53

 Feb 13, 2016

Best Answer 

 #3
avatar+130466 
+5

5x^2 + 3y^2 - 20x + 24y = -53       complete the square on x and y

 

5(x^2 - 4x  + 4)  +  3(y^2 + 8y + 16)  =   -53 + 20 + 48

 

5(x - 2)^2    +  3 (y + 4)^2  =   15     divide both sides by 15

 

(x - 2)^2 / 3   +  (y + 4)^2 / 5     =  1

 

This is an ellipse centered at ( 2 , -4)    with a major axis length  of 2sqrt(5)  and a minor axis length of 2sqrt(3)

 

Here's the graph : https://www.desmos.com/calculator/kklcz1gncn

 

 

cool cool cool

 Feb 13, 2016
 #1
avatar
0

5x (to the second power) + 3y (to the second power) - 20x + 24y = -53

 

I think!:

5x^2 +3y^2 - 20x + 24y +53 =0

 Feb 13, 2016
 #2
avatar+2499 
0

(xa)2+(yb)2=r2(equation of circle )5x2+3y220x+24y=535x2+3y220x+24y+(2)+(4)=53+(2)+(4)5x220x+2+3y2+24y+4=47see (a+b)2=a2+2ab+b2and (ab)2=a22ab+b2(5x2)2+(3x+2)2=53

 

your equation is WRONG may be you forget smth but i did it anyway just to show you, do the same on your equation 

 Feb 13, 2016
 #3
avatar+130466 
+5
Best Answer

5x^2 + 3y^2 - 20x + 24y = -53       complete the square on x and y

 

5(x^2 - 4x  + 4)  +  3(y^2 + 8y + 16)  =   -53 + 20 + 48

 

5(x - 2)^2    +  3 (y + 4)^2  =   15     divide both sides by 15

 

(x - 2)^2 / 3   +  (y + 4)^2 / 5     =  1

 

This is an ellipse centered at ( 2 , -4)    with a major axis length  of 2sqrt(5)  and a minor axis length of 2sqrt(3)

 

Here's the graph : https://www.desmos.com/calculator/kklcz1gncn

 

 

cool cool cool

CPhill Feb 13, 2016
 #4
avatar+2499 
+5

yea :)

Solveit  Feb 13, 2016

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