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Exercise 10.3 · Q11

Q.Find the equation for the ellipse that satisfies the given conditions: Vertices (0,±13)(0, \pm 13), foci (0,±5)(0, \pm 5).

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The vertices and foci lie on the yy-axis, so the ellipse has a vertical major axis with center at the origin. Using a=13a = 13 and c=5c = 5, we find b2=144b^2 = 144, giving the equation x2144+y2169=1\frac{x^2}{144} + \frac{y^2}{169} = 1.

Understanding the Standard Form

An ellipse is the set of all points whose distances from two fixed points (the foci) sum to a constant. When the ellipse is centered at the origin, its equation takes one of two standard forms depending on which axis is longer.

The key is to identify which axis contains the major diameter. The vertices are the endpoints of the major axis, and they're always farther from the center than any other point on the ellipse. The foci always lie on the major axis, between the center and the vertices.

Since both vertices (0,±13)(0, \pm 13) and foci (0,±5)(0, \pm 5) lie on the yy-axis, the major axis is vertical. For a vertical major axis, the standard form is:

x2b2+y2a2=1\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1

where a>b>0a > b > 0, and the relationship c2=a2−b2c^2 = a^2 - b^2 connects the semi-major axis length aa, semi-minor axis length bb, and focal distance cc.

For an ellipse with vertical major axis centered at the origin:

x2b2+y2a2=1wherec2=a2−b2\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1 \quad \text{where} \quad c^2 = a^2 - b^2

Finding the Equation

1. Identify aa from the vertices

The vertices are at (0,±13)(0, \pm 13), which means they're 1313 units from the center along the yy-axis. This distance is the semi-major axis length:

a=13a = 13

2. Identify cc from the foci …

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