How do you find the standard form of #y^2 - 5x^2 + 20x = 50# and what kind of a conic is it?

1 Answer

Answer:

Hyperbola #y^2/30-(x-2)^2/6=1#

Explanation:

From the given equation #y^2-5x^2+20x=50#

#y^2=5x^2-20x+50#

We complete the square
#y^2=5(x^2-4x)+50#

#y^2=5(x^2-4x+4-4)+50#

#y^2=5((x-2)^2-4)+50#

We return the 5 back inside the grouping symbol

#y^2=5(x-2)^2-20+50#

#y^2=5(x-2)^2+30#

transpose now all terms with variables to the left of the equation

#y^2-5(x-2)^2=30#

Divide both sides of the equation by 30 to make the right side 1

#y^2-5(x-2)^2=30#

#y^2/30-(5(x-2)^2)/30=30/30#

#y^2/30-(x-2)^2/6=1#

We can write this in a manner similar to #(y-k)^2/b^2-(x-h)^2/a^2=1#

#y^2/30-(x-2)^2/6=1#

#y^2/(sqrt(30))^2-(x-2)^2/(sqrt6)^2=1#

With asymptotes #y=sqrt5(x-2)# and #y=-sqrt5(x-2)#

graph{(y^2/30-(x-2)^2/6-1)=0[-40,40,-20,20]}

God bless...I hope the explanation is useful.