How do you determine circle, parabola, ellipse, or hyperbola from equation 4x^2 + 9y^2 - 16x +18y -11 = 0?

Jul 2, 2018

See below

Explanation:

$4 {x}^{2} + 9 {y}^{2} - 16 x + 18 y - 11 = 0$

Here's an easy way:
-If the coefficients on ${x}^{2}$ and ${y}^{2}$ match, it is a circle

-If there is only one squared term, it is a parabola

-If one of the squared terms has a negative coefficient, it is a hyperbola

-If the coefficients on ${x}^{2}$ and ${y}^{2}$ don't match but they still have coefficients that either both positive or both negative, it is a ellipse

This is an ellipse, let's put in it's standard form:
$4 {x}^{2} - 16 x + 9 {y}^{2} + 18 y = 11$

$4 \left({x}^{2} - 4 x + 4\right) + 9 \left({y}^{2} + 2 y + 1\right) = 11 + 16 + 9$

$4 {\left(x - 2\right)}^{2} + 9 {\left(y + 1\right)}^{2} = 36$

${\left(x - 2\right)}^{2} / 9 + {\left(y + 1\right)}^{2} / 4 = 1$

This is a horizontal ellipse

Ellipse

Explanation:

Compare the given general quadratic equation: $4 {x}^{2} + 9 {y}^{2} - 16 x + 18 y - 11 = 0$ with the standard form :$a {x}^{2} + 2 h x y + b {y}^{2} + 2 g x + 2 f y + c = 0$ we get

$a = 4 , h = 0 , b = 9 , g = - 8 , f = 9 , c = - 11$

Now, the determinant $\setminus \Delta$ of quadratic equation is given as follows

$\setminus \Delta = a b c + 2 f g h - a {f}^{2} - b {g}^{2} - c {h}^{2}$

$\setminus \Delta = 4 \setminus \cdot 9 \setminus \cdot \left(- 11\right) + 2 \setminus \cdot 9 \left(- 8\right) \left(0\right) - 4 {\left(9\right)}^{2} - 9 {\left(- 8\right)}^{2} - \left(- 11\right) {\left(0\right)}^{2}$

$\setminus \Delta = - 1296 \setminus \ne 0$

hence, the given quadratic equation: $4 {x}^{2} + 9 {y}^{2} - 16 x + 18 y - 11 = 0$ represents a conic section

Now, using second determinant

${h}^{2} - a b = {0}^{2} - \left(4\right) \left(9\right) = - 36 < 0$

Given quadratic equation $4 {x}^{2} + 9 {y}^{2} - 16 x + 18 y - 11 = 0$ represents an ellipse which can be re-written as follows

$4 {x}^{2} + 9 {y}^{2} - 16 x + 18 y - 11 = 0$

$4 \left({x}^{2} - 4 x + 4\right) + 9 \left({y}^{2} + 2 y + 1\right) - 25 - 11 = 0$

$4 {\left(x - 2\right)}^{2} + 9 {\left(y + 1\right)}^{2} = 36$

$\setminus \frac{4 {\left(x - 2\right)}^{2}}{36} + \setminus \frac{9 {\left(y + 1\right)}^{2}}{36} = 1$

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

Above ellipse has center at $\left(2 , - 1\right)$ & semi-major & semi-minor axes as $3$ & $2$ respectively