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

Concept: For the ellipse \[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1, \] the directrices are \[ x=\pm \frac{a}{e}. \] Thus, the given directrix immediately provides a relation between \(a\) and \(e\).

Step 1: Write the directrix in standard form. Given \[ \sqrt3\,x-4=0. \] Therefore, \[ x=\frac4{\sqrt3}. \] For the ellipse, \[ \frac{a}{e}=\frac4{\sqrt3}. \]

Step 2: Substitute the eccentricity. Given \[ e=\frac{\sqrt3}{2}. \] Hence \[ a = e\left(\frac4{\sqrt3}\right). \] \[ a = \frac{\sqrt3}{2} \cdot \frac4{\sqrt3}. \] \[ a=2. \]

Step 3: Find \(b\). Using \[ e^2=1-\frac{b^2}{a^2}, \] we get \[ \frac34 = 1-\frac{b^2}{4}. \] \[ \frac{b^2}{4} = \frac14. \] \[ b^2=1. \] \[ b=1. \]

Step 4: Compute \(ab\). \[ ab=(2)(1)=2. \] Since \[ a=2, \] we obtain \[ ab=a. \] \[ \boxed{ab=a}. \]