# How do use the method of translation of axes to sketch the curve x^2+4y^2-4x+24y+36=0?

Oct 7, 2017

The new axes are $\left(\begin{matrix}X \\ Y\end{matrix}\right) = \left(\begin{matrix}x - 2 \\ y + 3\end{matrix}\right)$

#### Explanation:

Let rewrite and factorise the equation of the curve

${x}^{2} + 4 {y}^{2} - 4 x + 24 y + 36 = 0$

${x}^{2} - 4 x + 4 \left({y}^{2} + 6 y\right) = - 36$

Complete the squares for $x$ and $y$

${x}^{2} - 4 x + {\left(- \frac{4}{2}\right)}^{2} + 4 \left({y}^{2} + 6 y + {\left(\frac{6}{2}\right)}^{2}\right) = - 36 + {\left(- \frac{4}{2}\right)}^{2} + {\left(\frac{6}{2}\right)}^{2}$

${x}^{2} - 4 x + 4 + 4 \left({y}^{2} + 6 y + 9\right) = - 36 + 4 + 36 = 4$

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

Dividing by $4$

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

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

Let the new system of axes be $\left(X , Y\right)$ such that

$X = x - 2$

$Y = y + 3$

Therefore,

${X}^{2} / 4 + {Y}^{2} / 1 = 1$

This is the equation of an ellipse.

graph{(x^2/4+y^2-1)((x-2)^2/4+(y+3)^2-1)=0 [-10, 10, -5, 5]}