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Chapter 6: Problem 29

Find the vertex, focus, and directrix of the parabola. Then sketch the parabola. $$y=\frac{1}{2} x^{2}$$

Short Answer

Expert verified
The vertex of the parabola \(y=0.5x^{2}\) is at the origin \((0,0)\). The focus of the parabola is at \((0,1)\) and the directrix of the parabola is \(y=-1\). The parabola opens upwards, with the focus above the vertex, and the directrix a horizontal line below the vertex.

Step by step solution

01

Identify the Vertex

Comparing our given equation with the standard equation \(y=ax^2\), we can see that there has been no vertical or horizontal shift indicated in our equation. Thus, our vertex is located at the origin \((0,0)\).
02

Find the Focus

The formula for the focus of a parabola is \(f=(h, k+1/(4a))\). Substituting the values of \(a=0.5\), \(h=0\) and \(k=0\) into the formula, we have \(f=(0, 0+1/(2*0.5))\), which simplifies to \(f=(0, 1)\). Therefore, the focus of the parabola is at \((0,1)\).
03

Find the Directrix

The equation of the directrix of a parabola is given as \(y=k-1/(4a)\). Substituting the values of \(a=0.5\), \(h=0\) and \(k=0\) into the formula, we have \(y = 0 - 1/(2*0.5)\), which simplifies to \(y=-1\). Therefore, the equation of the directrix of the parabola is \(y=-1\).
04

Sketch the Parabola

The final step of this exercise is to sketch the parabola based on the vertex, focus, and directrix that have been identified. The vertex is at the origin \((0,0)\). The focus is one unit above the vertex at \((0,1)\), and the directrix is one unit below the vertex, or \(y=-1\). Draw the x and y axes, plot these points and draw the parabola curve, ensuring it 'opens' towards the focus and 'away' from the directrix.

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