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Chapter 7: Problem 71

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

Short Answer

Expert verified
The standard form of the ellipse equation is \(\frac{{5x^2}}{{24}} + \frac{{y^2}}{{36}} = 1\)

Step by step solution

01

Identify Major Axis Length

The major axis is the line that passes through the foci of the ellipse. Vertices (\(0,-6\)) and (\(0,6\)) are on the same vertical line, so we know the major axis is vertical. The distance between them is the length of the major axis. Using the distance formula, which is \(d= \sqrt{{(x_2-x_1)^2 + (y_2-y_1)^2}}\), gives us \(d= \sqrt{{(0-0)^2 + (6 - (-6))^2}}\) = 12. The standard form of an ellipse formula is \(\frac{{x^2}}{{a^2}} + \frac{{y^2}}{{b^2}} = 1\). As the ellipse is vertically oriented, we have \(a= 6\) (half of the major axis length) and the denominators are switched, so \(y\) corresponds to \(a=6\) and \(x\) to \(b\).
02

Find the Length of Minor Axis

The minor axis length can be achieved through using the provided point \((2, -4)\). Substitute these \(x\) and \(y\) values into the standard ellipse equation and \(a = 6\), then solve for \(b\). The equation becomes \(\frac{{2^2}}{{b^2}} + \frac{{(-4)^2}}{{6^2}} = 1\). After simplifying, \(\frac{{4}}{{b^2}} + \frac{{16}}{{36}} = 1\). This leads to the equation \(\frac{{4}}{{b^2}} = \frac{{5}}{{6}}\), from which we can isolate \(b\).
03

Calculate Value of b and Construct Ellipse Equation

By cross multiplying and then square rooting, we find \(b = \sqrt{{\frac{{24}}{{5}}}} = 2 \sqrt{{6/5}}\). After obtaining the exact value, plug in \(a = 6\) and \(b = 2 \sqrt{{6/5}}\) into the ellipse standard equation and we get \(\frac{{x^2}}{{(2\sqrt{{6/5}})^2}} + \frac{{y^2}}{6^2} = 1\). This simplifies to \(\frac{{5x^2}}{{24}} + \frac{{y^2}}{{36}} = 1\), which is the standard form of the requested ellipse equation.

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Most popular questions from this chapter

Use a graphing utility to graph the parabolas.Write the given equation as a quadratic equation in \(y\) and use the quadratic formula to solve for \(y .\) Enter each of the equations to produce the complete graph. $$y^{2}+10 y-x+25=0$$

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

Find the standard form of the equation of the hyperbola with vertices \((5,-6)\) and \((5,6),\) passing through \((0,9)\).

Graph each ellipse and give the location of its foci. $$\frac{(x+2)^{2}}{16}+(y-3)^{2}=1$$

In Exercises \(1-18,\) graph each ellipse and locate the foci. $$4 x^{2}+16 y^{2}=64$$
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