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Chapter 7: Problem 23

Find the standard form of the equation of each parabola satisfying the given conditions. Focus: \((0,-25) ;\) Directrix: \(y=25\)

Short Answer

Expert verified
The standard form of the equation of the parabola is \(y = (1/100)x^2\).

Step by step solution

01

Find the Vertex of the Parabola

The vertex of the parabola is located exactly between the focus and directrix. Since the focus is at \((0,-25)\) and the directrix is \(y=25\), the y-coordinate of the vertex (\(k\)) is halfway between -25 and 25, which is 0. Also since the x-coordinate of the focus is 0, and the vertex is aligned with the focus in case of a vertical parabola, the x-coordinate of the vertex (\(h\)) is also zero. Therefore, the vertex coordinates are (0,0).
02

Determine the Value of 'a'

We can find 'a' using the formula \(a = 1/(4f)\), where 'f' is the distance of the vertex from the focus. Since our vertex is (0,0) and the focus is (0, -25), 'f' equals 25. Plugging this into our formula gives \(a = 1/(4*25) = 1/100\).
03

Write the Standard Form of the Equation

The standard form for this parabola is \(y = a(x - h)^2 + k\). We found that \(a = 1/100\), \(h = 0\), and \(k = 0\). Substituting these into the formula, we find the standard form of the equation of the parabola to be \(y = (1/100)x^2\)

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Most popular questions from this chapter

Find the standard form of the equation of an ellipse with vertices at \((0,-6)\) and \((0,6),\) passing through \((2,-4)\)

Convert each equation to standard form by completing the square on \(x\) and \(y .\) Then graph the hyperbola. Locate the foci and find the equations of the asymptotes. $$4 x^{2}-9 y^{2}-16 x+54 y-101=0$$

Graph \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) and \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=-1\) in the same viewing rectangle for values of \(a^{2}\) and \(b^{2}\) of your choice. Describe the relationship between the two graphs.

Find the standard form of the equation of each ellipse satisfying the given conditions. Foci: \((-2,0),(2,0) ; y\) -intercepts: \(-3\) and 3

In Exercises \(1-18,\) graph each ellipse and locate the foci. $$6 x^{2}=30-5 y^{2}$$
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