Question

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

Answer 1

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.