# How do you find the standard form given 4x^2-5y^2-40x-20y+160=0?

Sep 4, 2017

Given: $4 {x}^{2} - 5 {y}^{2} - 40 x - 20 y + 160 = 0$

Subtract 160 from both sides:

$4 {x}^{2} - 5 {y}^{2} - 40 x - 20 y = - 160$

Add $4 {h}^{2}$ to both sides:

$4 {x}^{2} - 40 x + 4 {h}^{2} - 5 {y}^{2} - 20 y = - 160 + 4 {h}^{2}$

Remove the common factor, 4, from the first 3 terms:

$4 \left({x}^{2} - 10 x + {h}^{2}\right) - 5 {y}^{2} - 20 y = - 160 + 4 {h}^{2}$

Subtract $5 {k}^{2}$ from both sides:

$4 \left({x}^{2} - 10 x + {h}^{2}\right) - 5 {y}^{2} - 20 y - 5 {k}^{2} = - 160 + 4 {h}^{2} - 5 {k}^{2}$

Remove a common factor of -5 from the next 3 terms:

$4 \left({x}^{2} - 10 x + {h}^{2}\right) - 5 \left({y}^{2} + 4 y + {k}^{2}\right) = - 160 + 4 {h}^{2} - 5 {k}^{2} \text{ }$

Because we are looking for the square ${\left(x - h\right)}^{2} = {x}^{2} - 2 h x + {h}^{2}$ we use the equation:

$- 2 h x = - 10 x$

To find the value of h:

$h = 5$

Into equation , substitute ${\left(x - 5\right)}^{2} \text{ for } {x}^{2} - 10 x + {h}^{2}$ and 100 for $4 {h}^{2}$:

$4 {\left(x - 5\right)}^{2} - 5 \left({y}^{2} + 4 y + {k}^{2}\right) = - 160 + 100 - 5 {k}^{2} \text{ [1.1]}$

Because we are looking for the square ${\left(y - k\right)}^{2} = {y}^{2} - 2 k y + {k}^{2}$ we use the equation:

$- 2 k y = 4 y$

To find the value of k:

$k = - 2$

Into equation [1.1], substitute ${\left(y - \left(- 2\right)\right)}^{2} \text{ for } {y}^{2} + 4 y + {k}^{2}$ and -20 for $- 5 {k}^{2}$:

$4 {\left(x - 5\right)}^{2} - 5 {\left(y - \left(- 2\right)\right)}^{2} = - 160 + 100 - 20 \text{ [1.2]}$

Combine like terms on the right:

$4 {\left(x - 5\right)}^{2} - 5 {\left(y - \left(- 2\right)\right)}^{2} = - 80 \text{ [1.3]}$

Divide both sides by -80:

${\left(y - \left(- 2\right)\right)}^{2} / 16 - {\left(x - 5\right)}^{2} / 20 = 1 \text{ [1.3]}$

Write the denominators as squares:

${\left(y - \left(- 2\right)\right)}^{2} / {\left(4\right)}^{2} - {\left(x - 5\right)}^{2} / {\left(2 \sqrt{5}\right)}^{2} = 1 \text{ [1.4]}$

Equation [1.4] is standard form.