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

Hyperbola ${y}^{2} / 30 - {\left(x - 2\right)}^{2} / 6 = 1$

#### Explanation:

From the given equation ${y}^{2} - 5 {x}^{2} + 20 x = 50$

${y}^{2} = 5 {x}^{2} - 20 x + 50$

We complete the square
${y}^{2} = 5 \left({x}^{2} - 4 x\right) + 50$

${y}^{2} = 5 \left({x}^{2} - 4 x + 4 - 4\right) + 50$

${y}^{2} = 5 \left({\left(x - 2\right)}^{2} - 4\right) + 50$

We return the 5 back inside the grouping symbol

${y}^{2} = 5 {\left(x - 2\right)}^{2} - 20 + 50$

${y}^{2} = 5 {\left(x - 2\right)}^{2} + 30$

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

${y}^{2} - 5 {\left(x - 2\right)}^{2} = 30$

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

${y}^{2} - 5 {\left(x - 2\right)}^{2} = 30$

${y}^{2} / 30 - \frac{5 {\left(x - 2\right)}^{2}}{30} = \frac{30}{30}$

${y}^{2} / 30 - {\left(x - 2\right)}^{2} / 6 = 1$

We can write this in a manner similar to ${\left(y - k\right)}^{2} / {b}^{2} - {\left(x - h\right)}^{2} / {a}^{2} = 1$

${y}^{2} / 30 - {\left(x - 2\right)}^{2} / 6 = 1$

${y}^{2} / {\left(\sqrt{30}\right)}^{2} - {\left(x - 2\right)}^{2} / {\left(\sqrt{6}\right)}^{2} = 1$

With asymptotes $y = \sqrt{5} \left(x - 2\right)$ and $y = - \sqrt{5} \left(x - 2\right)$

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

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