# How do use the method of translation of axes to sketch the curve 16x^2+9y^2-320x-108y+1780=0?

Dec 10, 2017

Please refer to the Discussion in the Explanation.

#### Explanation:

The eqn. of the curve is

$S : 16 {x}^{2} + 9 {y}^{2} - 320 x - 108 y + 1780 = 0.$

$\therefore 16 {x}^{2} - 320 x + 9 {y}^{2} - 108 y = - 1780.$

$\therefore 16 \left({x}^{2} - 20 x\right) + 9 \left({y}^{2} - 12 y\right) = - 1780.$

Completing the squares, we have,

$16 \left\{\left({x}^{2} - 20 x + 100\right) - 100\right\} + 9 \left\{\left({y}^{2} - 12 y + 36\right) - 36\right\} = - 1780.$

$\therefore 16 \left\{{\left(x - 10\right)}^{2} - 100\right\} + 9 \left\{{\left(y - 6\right)}^{2} - 36\right\} = - 1780.$

$\therefore 16 {\left(x - 10\right)}^{2} - 1600 + 9 {\left(y - 6\right)}^{2} - 324 = - 1780.$

$\therefore 16 {\left(x - 10\right)}^{2} + 9 {\left(y - 6\right)}^{2} = 1600 + 324 - 1780 = 144.$

Dividing by $144 ,$ we have,

$S : {\left(x - 10\right)}^{2} / 9 + {\left(y - 6\right)}^{2} / 16 = 1.$

Now, shifting the Origin to the point $\left(10 , 6\right) ,$ suppose that,

$\left(x , y\right)$ becomes $\left(X , Y\right) .$

$\therefore x = X + 10 , y = Y + 6.$

$\Rightarrow \left(x - 10\right) = X , \mathmr{and} \left(y - 6\right) = Y .$

$\therefore S : {X}^{2} / 9 + {Y}^{2} / 16 = 1.$

This represents an ellipse with centre at $\left(0 , 0\right)$ in $\left(X , Y\right)$

system and having lengths of major and minor axes $8 \mathmr{and} 6 ,$

resp.