Question

The length of the latus rectum of the ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) \( (a>b) \) is \( \frac{8}{3} \). If the distance from the center of the ellipse to its focus is \( \sqrt{5} \), then \( \sqrt{a^2 + 6ab + b^2} = \text{?} \)

• A. 7
• B. \( \sqrt{12} \)
• C. \( \sqrt{3} \)
• D. 11 \bigskip

Hint:

To solve for \( a \) and \( b \) in problems involving ellipses, use the formula for the latus rectum and the relationship between \( a \) and \( b \) based on the distance to the focus. After determining \( a \) and \( b \), use them to calculate the required expression.

Answer 1

The Correct Option is A

We are given the equation of the ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), where \( a>b \). The length of the latus rectum is given as \( \frac{8}{3} \), and the distance from the center of the ellipse to its focus is \( \sqrt{5} \). Step 1: Use the formula for the length of the latus rectum The length of the latus rectum of an ellipse is given by the formula: \[ L = \frac{2b^2}{a} \] We are told that the length of the latus rectum is \( \frac{8}{3} \). Therefore, we have the equation: \[ \frac{2b^2}{a} = \frac{8}{3} \] Solving for \( b^2 \), we get: \[ b^2 = \frac{8a}{6} = \frac{4a}{3} \] Step 2: Use the relationship between \( a^2 \) and \( b^2 \) The distance from the center of the ellipse to its focus is given by \( \sqrt{a^2 - b^2} \), and we are told that this distance is \( \sqrt{5} \). Therefore, we have the equation: \[ \sqrt{a^2 - b^2} = \sqrt{5} \] Squaring both sides: \[ a^2 - b^2 = 5 \] Substitute \( b^2 = \frac{4a}{3} \) into this equation: \[ a^2 - \frac{4a}{3} = 5 \] Multiply through by 3 to eliminate the fraction: \[ 3a^2 - 4a = 15 \] Rearranging this equation: \[ 3a^2 - 4a - 15 = 0 \] Step 3: Solve the quadratic equation We can solve the quadratic equation \( 3a^2 - 4a - 15 = 0 \) using the quadratic formula: \[ a = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 3 \cdot (-15)}}{2 \cdot 3} \] \[ a = \frac{4 \pm \sqrt{16 + 180}}{6} \] \[ a = \frac{4 \pm \sqrt{196}}{6} \] \[ a = \frac{4 \pm 14}{6} \] Thus, we have two possible solutions for \( a \): \[ a = \frac{4 + 14}{6} = \frac{18}{6} = 3 \] or \[ a = \frac{4 - 14}{6} = \frac{-10}{6} = -\frac{5}{3} \] Since \( a>b \) and the value of \( a \) must be positive, we take \( a = 3 \). Step 4: Find \( b^2 \) Substitute \( a = 3 \) into the equation \( b^2 = \frac{4a}{3} \): \[ b^2 = \frac{4 \cdot 3}{3} = 4 \] So, \( b = 2 \). Step 5: Calculate \( \sqrt{a^2 + 6ab + b^2} \) We need to find \( \sqrt{a^2 + 6ab + b^2} \). Substitute \( a = 3 \) and \( b = 2 \): \[ \sqrt{a^2 + 6ab + b^2} = \sqrt{3^2 + 6 \cdot 3 \cdot 2 + 2^2} \] \[ = \sqrt{9 + 36 + 4} = \sqrt{49} = 7 \] Thus, the correct answer is \( 7 \). \bigskip