Question

The eccentricity of an ellipse, with its center at the origin is \( \frac{1}{3} \). If one of the directrices is \( x = 9 \), then the equation of ellipse is:

• A. \( 9x^2 + 8y^2 = 72 \)
• B. \( 8x^2 + 9y^2 = 72 \)
• C. \( 8x^2 + 7y^2 = 56 \)
• D. \( 7x^2 + 8y^2 = 56 \)

Hint:

For an ellipse, the two key equations you need are for the directrix (\( x = \pm a/e \)) and the relationship between \( a, b, e \) (\( b^2 = a^2(1-e^2) \)). Use the given information to solve for \( a \) and \( b^2 \), then substitute them into the standard ellipse equation \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \).

Answer 1

The Correct Option is B

We are given the eccentricity \( e = \frac{1}{3} \). The equation of a directrix for a horizontal ellipse centered at the origin is \( x = \frac{a}{e} \) or \( x = -\frac{a}{e} \). We are given one directrix as \( x = 9 \). So, \( \frac{a}{e} = 9 \). Substitute the value of \( e \): \[ \frac{a}{1/3} = 9 \implies 3a = 9 \implies a = 3 \] So, \( a^2 = 9 \). Now we need to find \( b^2 \). The relationship between \( a, b, \) and \( e \) for an ellipse is \( b^2 = a^2(1 - e^2) \). \[ b^2 = 9 \left(1 - \left(\frac{1}{3}\right)^2\right) \] \[ b^2 = 9 \left(1 - \frac{1}{9}\right) = 9 \left(\frac{8}{9}\right) = 8 \] Now we have \( a^2 = 9 \) and \( b^2 = 8 \). The equation of the ellipse is \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). \[ \frac{x^2}{9} + \frac{y^2}{8} = 1 \] To convert this to the form given in the options, find the least common multiple of the denominators, which is 72, and multiply the entire equation by it: \[ 72 \left(\frac{x^2}{9}\right) + 72 \left(\frac{y^2}{8}\right) = 72 \times 1 \] \[ 8x^2 + 9y^2 = 72 \] This matches option (B).