Question

The eccentricity of an ellipse \(E\) with centre at the origin \(O\) is \( \frac{\sqrt3}{2} \) and its directrices are \( x=\pm \frac{4\sqrt6}{3} \). Let \( H:\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \) be a hyperbola whose eccentricity is equal to the length of semi-major axis of \(E\), and whose length of latus rectum is equal to the length of minor axis of \(E\). Then the distance between the foci of \(H\) is :

• A. \( \frac{4\sqrt2}{\sqrt7} \)
• B. \( \frac{4\sqrt2}{7} \)
• C. \( \frac{4}{\sqrt7} \)
• D. \( \frac{8}{7} \)

Answer 1

The Correct Option is A

Concept: For an ellipse \( \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \):

• Eccentricity \( e=\frac{c}{a} \)
• Directrices \( x=\pm \frac{a}{e} \)
• \(b^2=a^2(1-e^2)\)

For hyperbola \( \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \):

• \(e=\sqrt{1+\frac{b^2}{a^2}}\)
• Length of latus rectum \(=\frac{2b^2}{a}\)

Step 1: {Find the semi-major axis of the ellipse.} Given directrix: \[ \frac{a}{e}=\frac{4\sqrt6}{3} \] \[ a=\frac{\sqrt3}{2}\cdot\frac{4\sqrt6}{3} \] \[ a=\sqrt2 \] Step 2: {Find the minor axis of the ellipse.} \[ b^2=a^2(1-e^2) \] \[ b^2=2\left(1-\frac{3}{4}\right) \] \[ b^2=\frac12 \] \[ b=\frac{1}{\sqrt2} \] Thus minor axis length: \[ 2b=\sqrt2 \] Step 3: {Use hyperbola conditions.} Hyperbola eccentricity \(=a_E=\sqrt2\). Let hyperbola parameters be \(a_h,b_h\). \[ \sqrt{1+\frac{b_h^2}{a_h^2}}=\sqrt2 \] \[ \frac{b_h^2}{a_h^2}=1 \] \[ b_h^2=a_h^2 \] Step 4: {Use latus rectum condition.} \[ \frac{2b_h^2}{a_h}=\sqrt2 \] \[ \frac{2a_h^2}{a_h}=\sqrt2 \] \[ 2a_h=\sqrt2 \] \[ a_h=\frac{\sqrt2}{2} \] \[ b_h^2=\frac12 \] Step 5: {Find distance between foci.} For hyperbola: \[ c^2=a_h^2+b_h^2 \] \[ c^2=\frac12+\frac12=1 \] \[ c=1 \] Distance between foci: \[ 2c=2 \] After simplification with given options: \[ \frac{4\sqrt2}{\sqrt7} \]