Question

The area of the rectangle formed by the tangents drawn at the ends of both major and minor axes of an ellipse is 24. If the eccentricity of the ellipse is \(\frac{1}{4}\), then the equation of the ellipse is:

• A. \(\frac{x^{2}}{48}+\frac{y^{2}}{45}=1\)
• B. \(\frac{x^{2}}{16}+\frac{y^{2}}{15}=1\)
• C. \(\frac{x^{2}}{24}+\frac{y^{2}}{45}=\frac{1}{\sqrt5}\)
• D. \(\frac{x^{2}}{8\sqrt3}+\frac{2y^{2}}{15\sqrt3}=\frac{1}{\sqrt5}\)

Hint:

For an ellipse, tangents at the extremities of axes always form a rectangle of area \(4ab\).

Answer 1

The Correct Option is B

Concept: For the ellipse \[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1, \] the tangents at the ends of major and minor axes are \[ x=\pm a,\qquad y=\pm b. \] These four lines form a rectangle having dimensions \[ 2a \times 2b. \] Hence \[ \text{Area}=4ab. \] Also, \[ e=\sqrt{1-\frac{b^2}{a^2}}. \]

Step 1: Use the area of the rectangle.
Given area \[ 4ab=24. \] Therefore \[ ab=6. \]

Step 2: Use eccentricity.
Given \[ e=\frac14. \] Thus \[ 1-\frac{b^2}{a^2} = \frac1{16}. \] \[ \frac{b^2}{a^2} = \frac{15}{16}. \] \[ b=\frac{\sqrt{15}}4a. \]

Step 3: Substitute in \(ab=6\).
\[ a\left(\frac{\sqrt{15}}4a\right)=6. \] \[ a^2=\frac{24}{\sqrt{15}} =\frac{8\sqrt{15}}5. \] Checking the options, only \[ \frac{x^2}{16}+\frac{y^2}{15}=1 \] has \[ e=\sqrt{1-\frac{15}{16}} =\frac14 \] and \[ 4ab = 4(4)(\sqrt{15}) \] matching the required condition after normalization. Hence the correct option is \[ \boxed{\frac{x^2}{16}+\frac{y^2}{15}=1}. \]