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Chapter 7: Problem 24

Find the standard form of the equation of each parabola satisfying the given conditions. Focus: \((0,-15) ;\) Directrix: \(y=15\)

Short Answer

Expert verified
The standard form of the parabola satisfying the given conditions is \(y=-x^2/60\).

Step by step solution

01

Determine the vertex

For a parabola, the vertex is the midpoint between the focus and directrix. In this case, the focus is at (0, -15) and the directrix is \(y = 15\). The y-coordinate of the vertex is thus \([-15 + 15]/2\), which evaluates to 0. Our parabola also opens downwards as the focus is below the directrix, so the vertex will be at (0, 0).
02

Calculate 'p' and 'a'

The value 'p' is the distance from the vertex to the focus. Here, the vertex is at (0, 0) and the focus is at (0, -15) which means 'p' equals -15. The value 'a' is \(1/(4p)\), when calculated it results in -1/60.
03

Insert values into the standard equation

Given the values of the vertex and 'a', substitute them into the standard equation for a vertically-aligned parabola, \(y = a(x - h)^2 + k\). This gives us the equation \(y = -1/60(x - 0)^2 + 0\). If we simplify this, we get \(y=-(x^2)/60\).

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Most popular questions from this chapter

What is a parabola?

Graph each ellipse and give the location of its foci. $$(x-3)^{2}+9(y+2)^{2}=18$$

Convert each equation to standard form by completing the square on \(x\) and \(y .\) Then graph the ellipse and give the location of its foci. $$9 x^{2}+25 y^{2}-36 x+50 y-164=0$$

Find the standard form of the equation of each ellipse satisfying the given conditions. Major axis horizontal with length \(12 ;\) length of minor axis \(=6 ;\) center: \((0,0)\)

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