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An ellipse is a type of conic section that appears as an elongated circle. It comes from cutting through a cone at an angle, allowing two symmetric halves. The standard form of an ellipse equation is \[\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1,\]where

• \((h, k)\) is the center of the ellipse.
• \(a\) and \(b\) are the semi-major and semi-minor axis lengths, corresponding to the ellipse's longest and shortest diameters.

In this exercise, the transformed equation \[2(x-1)^2 + 3(y+2)^2 = 14\]represents an ellipse after simplification and division by 14. Its center is at \((1, -2)\), with semi-axis lengths \(\sqrt{7}\) for the x-axis and \(\sqrt{\frac{14}{3}}\) for the y-axis. Ellipses possess symmetry about their center, making them unique amongst conic sections.

Completing the square is a valuable algebraic method used to rewrite quadratic equations into a form that reveals important properties like the vertex of a parabola or the center of an ellipse. The process involves transforming an expression like \(ax^2 + bx + c\) into \(a(x-h)^2 + k\). Here's how it's done:

• Focus on the quadratic and linear terms: For \(2x^2 - 4x\), first factor out the coefficient 2 to simplify: \(2(x^2 - 2x)\).
• Find a value that completes the square inside the parentheses \((x^2 - 2x + ?)\): reformat it into a perfect square form \((x-1)^2 - 1\). The number 1 here is derived by taking half of the linear coefficient \((-2/2 = -1)\) and squaring it.
• Repeat the process similarly for the \(y\) terms.

Completing the square not only helps in identifying the nature of conic sections like ellipses but also simplifies solving them further down the road.

Intercepts are fundamental points where a graph meets the axes. These points provide a useful reference for sketching conic sections like ellipses. In this instance, intercepts are determined by setting one variable to zero and solving for the other.For x-intercepts, where \(y = 0\), we evaluate the equation:\[2x^2 - 4x + 2 = 0\]Using the quadratic formula reveals no real solutions, indicating no x-intercepts.For y-intercepts, where \(x = 0\), substitute into the original equation:\[3y^2 + 12y + 2 = 0\]Applying the quadratic formula yields two solutions: \(y = -0.175\) and \(y = -3.825\). These solutions represent the points at which the ellipse intersects the y-axis. Knowing intercepts assists in correctly placing the ellipse on the coordinate plane, confirming its orientation and exact location.