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Chapter 7: Problem 28

Find the standard form of the equation of each parabola satisfying the given conditions. Focus: \((2,4) ;\) Directrix: \(x=-4\)

Short Answer

Expert verified
The standard form of the equation for the parabola is \( (y-4)^2 = 12(x+1) \)

Step by step solution

01

Determine the vertex of the parabola

The vertex of the parabola is the midpoint of the focus and a point on the directrix directly across from the focus. Since the focus is (2,4) and the directrix is x=-4, we can take any point on the directrix (like (-4,4)) and find the midpoint. The vertex (h,k) will then be \(([2 +(-4)]/2 , [4+4]/2) = (-1,4)\)
02

Find the value of p

The value of p can be found by the formula \(p = h - x\), where x is any x-value on the directrix. For the directrix \( x = -4\), and for h from vertex coordinates, \(h = -1\), so, \(p = -1 - (-4) = 3\)
03

Construct the parabolic equation

Now we can plug in the values of h, k and p into the standard formula \((y-k)^2 = 4p(x-h)\), to get \((y-4)^2 = 4*3 (x-(-1))\), simplifying this results in \((y-4)^2 = 12(x+1)\)

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Most popular questions from this chapter

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.

Graph each ellipse and give the location of its foci. $$\frac{(x+3)^{2}}{9}+(y-2)^{2}=1$$

Use the center, vertices, and asymptotes to graph each hyperbola. Locate the foci and find the equations of the asymptotes. $$\frac{(y-2)^{2}}{36}-\frac{(x+1)^{2}}{49}=1$$

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. $$4 x^{2}+25 y^{2}-24 x+100 y+36=0$$

What is an ellipse?
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