Question

For an ellipse, distance between centre and focus is \(\sqrt7\) and semi-latus rectum is \(\frac{9}{4}\). Area of triangle formed by foci and one end of minor axis is

• A. \(\frac{\sqrt7}{4}\)
• B. \(9\sqrt7\)
• C. \(3\sqrt7\)
• D. \(\frac{\sqrt7}{2}\) \bigskip

Hint:

In ellipse geometry, focus–minor axis triangle area simplifies to \(cb\).

Answer 1

The Correct Option is C

Concept: For ellipse: \[ c=\sqrt7,\quad \frac{b^2}{a}=\frac{9}{4} \]

Step 1: Find parameters.
Using: \[ c^2=a^2-b^2 \] and \[ \frac{b^2}{a}=\frac94 \] Solving gives: \[ a=4,\quad b=\frac{3\sqrt7}{2} \]

Step 2: Area of triangle.
Triangle formed by foci and minor axis end: \[ \text{Area}=c \cdot b \] \[ =\sqrt7 \cdot 3 = 3\sqrt7 \] \[ \boxed{(C)} \]