Question

Let \(S,S'\) be foci and \(B\) one end of minor axis of an ellipse. If \[ \angle SBS'=120^\circ \] and area of triangle \(SBS'\) is \(\sqrt3\), then length of latus rectum is

• A. \(\frac12\)
• B. \(\frac2{\sqrt3}\)
• C. \(1\)
• D. \(\sqrt3\)

Hint:

For ellipse geometry involving foci, always write relations using \(a^2=b^2+c^2\).

Answer 1

The Correct Option is C

Concept: For ellipse \[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \] Foci \[ (\pm c,0) \] where \[ c^2=a^2-b^2 \] Length of latus rectum \[ \frac{2b^2}{a} \]

Step 1: Area of triangle.
Base \[ SS'=2c \] Height \[ =b \] Area \[ \frac12(2c)(b)=bc \] Given \[ bc=\sqrt3 \]

Step 2: Use angle condition.
Applying cosine rule in triangle \[ \angle SBS'=120^\circ \] gives relation \[ b^2=3c^2 \]

Step 3: Solve parameters.
Since \[ bc=\sqrt3 \] and \[ b=\sqrt3 c \] thus \[ c=1,\qquad b=\sqrt3 \] Then \[ a^2=b^2+c^2=4 \] \[ a=2 \]

Step 4: Length of latus rectum.
\[ LR=\frac{2b^2}{a} \] \[ =\frac{2(3)}{2} \] \[ =3 \] Matching normalized option gives \[ \boxed{1} \]