# How do you identity if the equation x^2+2y^2=2x+8 is a parabola, circle, ellipse, or hyperbola and how do you graph it?

Apr 9, 2017

${x}^{2} + 2 {y}^{2} - 2 x - 8 = 0$ is an ellipse.

#### Explanation:

Let the equation be of the type $A {x}^{2} + B x y + C {y}^{2} + D x + E y + F = 0$

then if

${B}^{2} - 4 A C = 0$ and $A = 0$ or $C = 0$, it is a parabola

${B}^{2} - 4 A C < 0$ and $A = C$, it is a circle

${B}^{2} - 4 A C < 0$ and $A \ne C$, it is an ellipse

${B}^{2} - 4 A C > 0$, it is a hyperbola

This equation is ${x}^{2} + 2 {y}^{2} - 2 x - 8 = 0$ and hence, $A = 1$, $B = 0$, $C = 2$ and ${B}^{2} - 4 A C = - 8$

As ${B}^{2} - 4 A C < 0$ and $A \ne C$, hence it is an ellipse. We can write it as

${x}^{2} - 2 x - 8 + 2 {y}^{2} = 0$

or ${x}^{2} - 2 x + 1 + 2 {y}^{2} = 9$

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

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

Hence, this is n ellipse with center at $\left(1 , 0\right)$, major axis along $x$-axis as $3 \times 2 = 6$ and minor axis along $y$-axis $2 \times \frac{3}{\sqrt{2}} = 3 \sqrt{2}$. Hence end points for major axis on $x$-axis are $\left(- 2 , 0\right)$ and $\left(4 , 0\right)$ and end points for minor axis are $\left(1 , \frac{3}{\sqrt{2}}\right)$ and $\left(1 , - \frac{3}{\sqrt{2}}\right)$

graph{y(x^2-2x-8+2y^2)(x-1)=0 [-4.21, 5.79, -2.46, 2.54]}