How do I graph $16 x^{2} - 9 y^{2} + 32 x + 18 y - 137 = 0$ $16x^2-9y^2+32x+18y-137=0$ algebraically?

Question

3057 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-09-26T18:16:16

Ellipse centered at (-1,1) and semi major axis of 4 unitsalong y axisand semi minor axis of 3 units.

Write the equation in standard form by completing the squares,

$16 x^{2} + 32 x - 9 y^{2} + 18 y - 137 = 0$ $16x^2 +32x -9y^2 +18y -137=0$

$16 (x^{2} + 2 x + 1) - 9 (y^{2} - 2 y + 1) - 16 + 9 - 137 = 0$ $16(x^2 +2x +1) -9(y^2 -2y +1) -16 +9-137=0$

$16 {(x + 1)}^{2} - 9 {(y - 1)}^{2} = 144$ $16(x+1)^2 -9(y-1)^2=144$

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

This is an ellipse with its centre at (-1,1), with semi-major axis of 4 units along y axis and semi-minor axis of 3 units along x axis

How do I graph 16x2−9y2+32x+18y−137=016x^2-9y^2+32x+18y-137=0 algebraically?