Question

Area of the quadrilateral formed by joining the extremities of the major axis and minor axis of the ellipse \[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \] is \(8\sqrt3\). If the distance between its foci is \(4\sqrt2\), then the eccentricity of the ellipse is

• A. \(\dfrac{1}{\sqrt3}\)
• B. \(\dfrac13\)
• C. \(\dfrac23\)
• D. \(\sqrt{\dfrac23}\)

Hint:

For the ellipse \[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1, \] the quadrilateral formed by the vertices \((\pm a,0)\) and \((0,\pm b)\) is a rhombus with area \[ 2ab. \] Combine this with \[ c^2=a^2-b^2 \] to find the eccentricity quickly.

Answer 1

The Correct Option is D

Concept: For the ellipse \[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1, \qquad a\gt b, \] the extremities of the major and minor axes are \[ (\pm a,0),\qquad (0,\pm b). \] Joining these points forms a rhombus whose diagonals are \[ 2a \quad \text{and} \quad 2b. \] Hence its area is \[ \frac12(2a)(2b)=2ab. \] Also, \[ c^2=a^2-b^2, \qquad e=\frac{c}{a}. \]

Step 1: Use the given area of the quadrilateral. Given \[ 2ab=8\sqrt3. \] Therefore, \[ ab=4\sqrt3. \] Squaring, \[ a^2b^2=48. \] \[ \cdots (1) \]

Step 2: Use the distance between the foci. Distance between the foci is \[ 2c=4\sqrt2. \] Hence, \[ c=2\sqrt2. \] \[ c^2=8. \] Since \[ c^2=a^2-b^2, \] we obtain \[ a^2-b^2=8. \] \[ \cdots (2) \]

Step 3: Find \(a^2\) and \(b^2\). Let \[ A=a^2, \qquad B=b^2. \] Then from (1) and (2), \[ AB=48, \] \[ A-B=8. \] Substituting \[ A=B+8, \] \[ B(B+8)=48. \] \[ B^2+8B-48=0. \] \[ (B-4)(B+12)=0. \] Since \(B\gt 0\), \[ B=4. \] Thus, \[ A=12. \] Hence, \[ a^2=12, \qquad b^2=4. \]

Step 4: Find the eccentricity. \[ e = \frac{c}{a} = \sqrt{\frac{c^2}{a^2}}. \] \[ = \sqrt{\frac{8}{12}}. \] \[ = \sqrt{\frac23}. \]

Step 5: Write the final answer. \[ \boxed{\sqrt{\frac23}} \]