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Problem 757

Prove that if \(\mathrm{F}(0,1)\) is the focus, and the line \(\mathrm{y}=-1\) is the directrix, then the equation of the parabola is $\mathrm{y}=(1 / 4) \mathrm{x}^{2}$.

Short Answer

Expert verified
To prove that the given parabola has the focus F(0,1) and directrix y = -1, we define a general point P(x,y) on the parabola and find the distances PF and PD between the point and the focus, and the point and the directrix, respectively. Using the definition of a parabola, we equate the two distances, square both sides, and simplify the equation to obtain \(x^2 = 4y\). Dividing by 4, we get the final equation of the parabola as \(y = \frac{1}{4}x^2\).
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Step 1: Define a general point on the parabola

Let P(x,y) be a general point on the parabola.

Step 2: Find the distance between the point and the focus

Using the distance formula: \(PF = \sqrt{(x-0)^2 + (y-1)^2} = \sqrt{x^2 + (y-1)^2}\).

Step 3: Find the distance between the point and the directrix

Since the directrix is a horizontal line at y=-1, the vertical distance between the point P(x,y) and the directrix is given by: \(PD = y - (-1) = y + 1\).

Step 4: Equate the two distances

According to the definition of a parabola, the distance between the point and the focus must be equal to the distance between the point and the directrix. Therefore, we have: \(\sqrt{x^2 + (y-1)^2} = y + 1\).

Step 5: Simplify the equation

To get rid of the square root, square both sides of the equation: \((x^2 + (y-1)^2) = (y + 1)^2\). Expand both sides: \(x^2 + y^2 - 2y + 1 = y^2 + 2y + 1\). Now, notice that y^2 and 1 terms cancel out and simplify the equation to obtain the equation of the parabola: \(x^2 = 4y\).

Step 6: Divide by 4 to get the final equation

To get the final equation of the parabola, divide both sides by 4: \(y = \frac{1}{4}x^2\). Thus, we have proved that if F(0,1) is the focus and the line y = -1 is the directrix, then the equation of the parabola is \(y = \frac{1}{4}x^2\).

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Most popular questions from this chapter

Chapter 43

Plot points of the curve corresponding to $\mathrm{y}=(1 / 4)(\mathrm{x}-2)^{2}\( for \)\mathrm{x}=4.3 .2 .1 .0 \mathrm{and}$ sketch the curve.

Chapter 43

Consider the parabola \(y^{2}=4 p x .\) A tangent to the parabola at point \(\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\) is defined as the line that intersects the parabola at point \(\mathrm{P}_{1}\) and nowhere else. (a) Show that the slope of the tangent line is \(2 \mathrm{p} / \mathrm{y}_{1}\) [Hint: Let the slope be \(\mathrm{m}\). Find the equation of the line passing through \(\mathrm{P}_{1}\) with slope \(\mathrm{m}\). What are the points of intersection of the tangent line and the parabola? For what values of \(\mathrm{m}\), would there be only one intersection point?] (b) Find the equation of the tangent line. (c) Prove that the intercepts of the tangent line are $\left(-\mathrm{x}_{1}, 0\right)\( and \)\left[0,(1 / 2) \mathrm{y}_{1}\right]$

Chapter 43

Consider the equation \(x^{2}-4 y^{2}+4 x+8 y+4=0\) Express this equation in standard form, and determine the center, the vertices, the foci, and the eccentricity of this hyperbola. Describe the fundamental rectangle and find the equations of the 2 asymptotes.

Chapter 43

Determine the intercepts, find the asymptotes, and locate the foci of the following hyperbolas: (a) \(x^{2}-\left(y^{2} / 4\right)=1\). (b) \(\left(y^{2} / 16\right)-\left(x^{2} / 4\right)=1\).

Chapter 43

Consider the equation of a parabola $\mathrm{x}^{2}-4 \mathrm{x}-4 \mathrm{y}+8=0$. Find the focus, vertex, axis of symmetry, and the directrix.

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