Question

If the eccentricity of the ellipse \[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \qquad (a>b) \] is \[ e=\frac{\sqrt3}{2} \] and the equation of one of its directrices is \[ \sqrt3\,x-4=0, \] then \(ab=\):

• A. \(a\)
• B. \(b\)
• C. \(a^2-b^2\)
• D. \(\dfrac{e}{a}\)

Hint:

For the ellipse \[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1, \] always remember: \[ e=\sqrt{1-\frac{b^2}{a^2}}, \] and the directrices are \[ x=\pm \frac{a}{e}. \] These two formulas alone solve most directrix–eccentricity problems.

Answer 1

The Correct Option is A