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Exercise 10.3 · Q2

Q.Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse x24+y225=1\frac{x^2}{4} + \frac{y^2}{25} = 1.

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The ellipse x24+y225=1\frac{x^2}{4} + \frac{y^2}{25} = 1 has its major axis along the yy-axis (since 25>425 > 4). Vertices at (0,±5)(0, \pm 5), foci at (0,±21)(0, \pm \sqrt{21}), major axis length 1010, minor axis length 44, eccentricity 215\frac{\sqrt{21}}{5}, and latus rectum 85\frac{8}{5}.

The standard form of an ellipse centered at the origin is x2a2+y2b2=1\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1. The key insight is to identify which denominator is larger—that determines which axis is major. Here 25>425 > 4, so the major axis lies along the yy-axis. This means b2=25b^2 = 25 (giving b=5b = 5) and a2=4a^2 = 4 (giving a=2a = 2).

For an ellipse with major axis along the yy-axis, the relationship between the semi-axes and the focal distance cc is:

c2=b2−a2c^2 = b^2 - a^2

This comes from the geometric definition: the sum of distances from any point on the ellipse to the two foci is constant and equals 2b2b (the length of the major axis).

Let me work through each element systematically.

1. Calculate the focal distance cc

c2=b2−a2=25−4=21c^2 = b^2 - a^2 = 25 - 4 = 21

So c=21c = \sqrt{21}.

2. Coordinates of the foci

Since the major axis is vertical, the foci lie on the yy-axis at (0,±c)(0, \pm c).

Foci: (0,21)(0, \sqrt{21}) and (0,−21)(0, -\sqrt{21})

3. Coordinates of the vertices

The vertices are the endpoints of the major axis, located at (0,±b)(0, \pm b).

Vertices: (0,5)(0, 5) and (0,−5)(0, -5)

4. Length of the major axis

The major axis spans from one vertex to the other, so its length is 2b2b.

Length of major axis: 2×5=102 \times 5 = 10

5. Length of the minor axis

The minor axis has length 2a2a.

Length of minor axis: 2×2=42 \times 2 = 4

6. Eccentricity ee

Eccentricity measures how "stretched" the ellipse is, defined as:

e=cbe = \frac{c}{b}

where bb is the semi-major axis length.

e=215e = \frac{\sqrt{21}}{5}

Tip

For an ellipse, 0<e<10 < e < 1 always. The closer ee is to 11, the more elongated the ellipse; closer to 00 means more circular.

7. Length of the latus rectum

The latus rectum is a chord through a focus, perpendicular to the major axis. Its length is given by:

Latus rectum=2a2b\text{Latus rectum} = \frac{2a^2}{b}

=2×45=85= \frac{2 \times 4}{5} = \frac{8}{5}

Watch out

A common mistake is to always assume a2a^2 is the larger denominator. Always compare the denominators first—the larger one corresponds to the major axis direction.

✓Final answer

The ellipse has foci at (0,±21)(0, \pm\sqrt{21}), vertices at (0,±5)(0, \pm 5), major axis length 1010, minor axis length 44, eccentricity 215\frac{\sqrt{21}}{5}, and latus rectum 85\frac{8}{5}.

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