Question

An ellipse has its centre at \((1,-2)\), one focus at \((3,-2)\) and one vertex at \((5,-2)\). Then the length of its latus rectum is:

• A. \(\dfrac{16}{\sqrt3}\)
• B. \(6\)
• C. \(4\sqrt3\)
• D. \(6\sqrt3\)

Hint:

When centre, focus, and vertex lie on the same horizontal line, the major axis is along the \(x\)-axis.

Answer 1

The Correct Option is B

To find the length of the latus rectum of the ellipse, we need to determine key parameters of the ellipse based on the information given.

The equation of an ellipse in standard form with center \((h, k)\) parallel to the coordinate axes is:

\[\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\]

where \(a\) is the semi-major axis and \(b\) is the semi-minor axis. For a horizontal ellipse, if \(c\) is the distance from the center to each focus, then:

• \(c^2 = a^2 - b^2\)
• Length of the latus rectum \(= \frac{2b^2}{a}\)

Based on the problem:

• Center \((h, k) = (1, -2)\)
• One focus \((h+c, k) = (3, -2)\), hence \(c = 2\)
• One vertex \((h+a, k) = (5, -2)\), hence \(a = 4\)

With \(a = 4\) and \(c = 2\), we use the relationship:

\(c^2 = a^2 - b^2\)

\(2^2 = 4^2 - b^2\)

\(4 = 16 - b^2\)

\(b^2 = 12\)

Now, calculate the length of the latus rectum:

\(\text{Length of Latus Rectum} = \frac{2b^2}{a} = \frac{2 \times 12}{4} = 6\)

This matches the correct answer.

Therefore, the length of the latus rectum of the ellipse is \(6\).