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Exercise 10.3 · Q15

Q.Find the equation for the ellipse that satisfies the given conditions: Length of major axis 2626, foci (±5,0)(\pm 5, 0).

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The ellipse has its major axis along the x-axis, with a=13a = 13 and c=5c = 5. Using b2=a2−c2b^2 = a^2 - c^2, we get b2=144b^2 = 144. The equation is x2169+y2144=1\frac{x^2}{169} + \frac{y^2}{144} = 1.

The key to any ellipse problem is understanding what the given numbers actually mean geometrically. The foci are at (±5,0)(\pm 5, 0), which tells you two things immediately: the ellipse is centered at the origin (since the foci are symmetric about the origin), and the major axis is along the x-axis (because the foci lie on the x-axis). The length of the major axis is 26, so the semi-major axis aa is half of that: a=13a = 13.

Now, for an ellipse centered at the origin with the major axis along the x-axis, the standard equation is:

x2a2+y2b2=1\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1

where a>b>0a > b > 0. The foci are at (±c,0)(\pm c, 0), and the relationship between aa, bb, and cc is:

c2=a2−b2c^2 = a^2 - b^2

You already have a=13a = 13 and c=5c = 5. The only missing piece is bb, and the formula above gives it directly.

  1. Identify aa from the major axis length.

    The length of the major axis is 2a=262a = 26, so a=13a = 13.

  2. Identify cc from the foci.

    The foci are (±5,0)(\pm 5, 0), so c=5c = 5.

  3. Use the ellipse relationship to find b2b^2.

b2=a2−c2=132−52=169−25=144b^2 = a^2 - c^2 = 13^2 - 5^2 = 169 - 25 = 144

Hence b=12b = 12 (since b>0b > 0).

  1. Write the equation. …

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