# How do I convert the equation 4x^2 - y^2 - 24x+4y +28= 0 to standard form?

Nov 15, 2015

See explanation

#### Explanation:

First, we group the terms with $x$ and those with $y$

$4 {x}^{2} - {y}^{2} - 24 x + 4 y + 28 = 0$

$\implies \left(4 {x}^{2} - 24 x\right) - \left({y}^{2} - 4 y\right) + 28 = 0$

Next, we "complete" the squares.
We do this by adding a third term such that the $x$ terms and the $y$ terms are perfect squares. However, when we do this, we must either add the same value on the other side of the equation or subtract the same value on the same side so that the equality is maintained

$\implies \left(4 {x}^{2} - 24 x\right) - \left({y}^{2} - 4 y\right) + 28 = 0$

$\implies 4 \left({x}^{2} - 6 x\right) - \left({y}^{2} - 4 y\right) + 28 = 0$

$\implies 4 \left({x}^{2} - 6 x + 9\right) - \left({y}^{2} - 4 y + 4\right) + 28 = 0 + 4 \left(9\right) - 1 \left(4\right)$

Then, we simplify.

$\implies 4 {\left(x - 3\right)}^{2} - {\left(y - 2\right)}^{2} + 28 = 32$

$\implies 4 {\left(x - 3\right)}^{2} - {\left(y - 2\right)}^{2} = 32 - 28$

$\implies 4 {\left(x - 3\right)}^{2} - {\left(y - 2\right)}^{2} = 4$

Finally, we divide both sides by a number such that the right-hand side will be 1

$\implies {\left(x - 3\right)}^{2} - {\left(y - 2\right)}^{2} / 4 = 1$