91Ó°ÊÓ

First, we rewrite the equation in the general conic form: \(Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0\). In our case, we have \(A = 2\), \(B = 0\), \(C = 2\), \(D = 0\), \(E = 0\), and \(F = -13\). To determine the type of conic section, we can use the discriminant, defined as: \(\Delta = B^2 - 4AC\). In our case, we have: \(\Delta = 0^2 - 4\cdot2\cdot2 = 0 - 16 = -16\). Since \(\Delta < 0\), we have an ellipse.

Now, we need to rewrite the equation in the standard form for ellipses. The standard form for ellipses is given by: \(\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1\), where \((h, k)\) is the center of the ellipse, and \(a\) and \(b\) are the lengths of the semi-major and semi-minor axes, respectively. Our given equation is already simplified, and we can rewrite it to obtain the standard form: \(2x^2 + 2y^2 = 13\) Now, divide both sides by 13: \(\frac{x^2}{(13/2)} + \frac{y^2}{(13/2)} = 1\) Comparing this equation to the standard form, we see that, in our case, \(h = 0\), \(k = 0\), \(a^2 = 13/2\), and \(b^2 = 13/2\). We obtain the values of \(a\) and \(b\): \(a = \sqrt{13/2}\) and \(b = \sqrt{13/2}\).

Now that we have all the values needed, we can sketch the ellipse using the following steps: 1. Plot the center of the ellipse at (0, 0). 2. Draw the major and minor axes with lengths \(2a\) and \(2b\), respectively. 3. Since the ellipse's semi-major and semi-minor axes have the same length, it is a circle. 4. Draw the circle centered at the origin with radius \(r = a = b = \sqrt{13/2}\). The graph of the equation \(2x^2 + 2y^2 - 13 = 0\) is a circle centered at the origin with radius \(\sqrt{13/2}\).