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

Concept: For an ellipse with major axis along the \(x\)-axis: \[ \text{Length of latus rectum} = \frac{2b^2}{a}, \] where \[ a=\text{semi-major axis},\quad b=\text{semi-minor axis},\quad c=\text{distance of focus from centre}. \] These quantities satisfy: \[ c^2 = a^2 - b^2. \]
Step 1: Identify orientation and parameters Since the centre, focus, and vertex all lie on the line \(y=-2\), the major axis is along the \(x\)-direction. Centre: \((1,-2)\) Focus: \((3,-2)\) \[ c = |3-1| = 2 \] Vertex: \((5,-2)\) \[ a = |5-1| = 4 \]
Step 2: Find \(b^2\) Using the relation: \[ c^2 = a^2 - b^2 \] \[ 2^2 = 4^2 - b^2 \Rightarrow 4 = 16 - b^2 \Rightarrow b^2 = 12 \]
Step 3: Compute the length of latus rectum \[ L = \frac{2b^2}{a} = \frac{2\times 12}{4} = 6 \] Final Answer: \[ \boxed{6} \]