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Chapter 6: Problem 29

Find the standard form of the equation of the parabola with the given characteristic(s) and vertex at the origin. Vertical axis and passes through the point (4,6)

Short Answer

Expert verified
The standard form of the equation of the parabola is \(y = 0.375x^2\).

Step by step solution

01

Identify the Type of Parabola

The exercise mentions that the parabola has a vertical axis, so the standard form of our equation must look like this: \(y = ax^2\). Since the vertex is at the origin, the equation does not contain \(h\) and \(k\). The variable 'a' determines the opening direction and the width of the parabola.
02

Substitute the Given Point

We need to substitute the given point (4,6) into the equation to solve it for 'a'. So we replace \(x = 4\) and \(y = 6\) in the equation getting \(6 = a * 4^2\).
03

Calculate the Variable 'a'

Solve the equation for 'a'. The calculation is as follows: \(6 = 16a\) then divide by 16 on both sides to isolate 'a', getting \(a = \frac{6}{16} = 0.375\).

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Most popular questions from this chapter

Sketch the graph of the ellipse, using latera recta. \(\frac{x^{2}}{4}+\frac{y^{2}}{1}=1\)

Find an equation of the tangent line to the parabola at the given point, and find the \(x\) -intercept of the line. \(x^{2}=2 y,(4,8)\)

Use a graphing utility to graph the ellipse. Find the center, foci, and vertices. (Recall that it may be necessary to solve the equation for \(y\) and obtain two equations.) \(3 x^{2}+4 y^{2}=12\)

Find the equation of an ellipse such that for any point on the ellipse, the sum of the distances from the point (2,2) and (10,2) is 36 .

Find an equation of the ellipse with vertices (卤5,0) and eccentricity \(e=\frac{3}{5}\).
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