Question

If the latus rectum of the ellipse \[ \frac{x^2}{16}+\frac{y^2}{b^2}=1 \] subtends an angle \[ \frac{\pi}{3} \] at the centre of the ellipse, then \(b^2=\)

• A. \(9\)
• B. \(\dfrac94(\sqrt{13}-1)\)
• C. \(\dfrac83(\sqrt{13}-1)\)
• D. \(7\)

Hint:

For an ellipse \[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1, \] the endpoints of the latus rectum are \[ \boxed{\left(ae,\pm\frac{b^2}{a}\right)}. \] Use the angle made by the position vectors from the centre.

Answer 1

The Correct Option is C

Step 1: Write the coordinates of the latus rectum. For the ellipse \[ \frac{x^2}{16}+\frac{y^2}{b^2}=1, \] we have \[ a=4. \] The endpoints of the right latus rectum are \[ \left(ae,\pm\frac{b^2}{a}\right). \] Hence the angle subtended at the centre is \[ 2\tan^{-1}\left(\frac{b^2}{a^2e}\right). \]

Step 2: Use the given angle. Given, \[ 2\tan^{-1}\left(\frac{b^2}{16e}\right) =\frac{\pi}{3}. \] Therefore, \[ \tan\frac{\pi}{6} = \frac{b^2}{16e}, \] i.e., \[ \frac1{\sqrt3} = \frac{b^2}{16e}. \] Hence, \[ b^2=\frac{16e}{\sqrt3}. \]

Step 3: Use the relation between \(b\) and \(e\). Since \[ e=\sqrt{1-\frac{b^2}{16}}, \] substituting into the above equation and solving, \[ 3b^4+16b^2-256=0. \] The positive solution is \[ b^2 = \frac83(\sqrt{13}-1). \] Therefore, \[ \boxed{\frac83(\sqrt{13}-1)}. \] Thus, \[ \boxed{(C)} \] is the correct answer.