Question

If the major axis of an ellipse is parallel to the \(X\)-axis and the ratio of the lengths of its latus rectum and minor axis is \(2:3\), then the ratio of distances from its centre to its focus and directrix is

• A. \(5:9\)
• B. \(4:5\)
• C. \(3:5\)
• D. \(5:8\)

Hint:

For an ellipse, \[ \boxed{\text{Latus rectum}=\frac{2b^2}{a}} \] and \[ \boxed{\text{Distance from centre to directrix}=\frac{a}{e}.} \] Also, \[ \boxed{\frac{\text{Centre to focus}}{\text{Centre to directrix}}=e^2.} \]

Answer 1

The Correct Option is A

Step 1: Recall the formulae for an ellipse. For the ellipse \[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1, \qquad (a>b), \] the length of the latus rectum is \[ \frac{2b^2}{a}, \] and the length of the minor axis is \[ 2b. \]

Step 2: Use the given ratio. Given, \[ \frac{\text{Length of latus rectum}}{\text{Length of minor axis}} = \frac23. \] Therefore, \[ \frac{\dfrac{2b^2}{a}}{2b} = \frac23. \] Simplifying, \[ \frac{b}{a} = \frac23. \] Hence, \[ b=\frac{2a}{3}. \]

Step 3: Find the eccentricity. Using \[ e=\sqrt{1-\frac{b^2}{a^2}}, \] we get \[ e = \sqrt{1-\frac49} = \sqrt{\frac59} = \frac{\sqrt5}{3}. \] Thus, \[ c=ae=\frac{a\sqrt5}{3}. \] The distance from the centre to the directrix is \[ \frac{a}{e} = \frac{a}{\sqrt5/3} = \frac{3a}{\sqrt5}. \]

Step 4: Find the required ratio. Hence, \[ \frac{\text{Centre to focus}}{\text{Centre to directrix}} = \frac{c}{a/e} = \frac{ae}{a/e} = e^2 = \frac59. \] Therefore, \[ \boxed{5:9.} \] Hence, the correct option is \(\boxed{(A)}\).