Question

Let the length of a latus rectum of an ellipse $ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $ be 10. If its eccentricity is $ e $, and the minimum value of the function $ f(t) = t^2 + t + \frac{11}{12} $, where $ t \in \mathbb{R} $, then $ a^2 + b^2 $ is equal to:

• A. 125
• B. 126
• C. 120
• D. 115

Hint:

To solve problems involving the latus rectum and eccentricity of an ellipse, remember the relationships between \( a^2 \), \( b^2 \), and the length of the latus rectum. Use optimization to minimize given functions and relate them to the geometry of the ellipse.

Answer 1

The Correct Option is B

To solve this problem, we need to determine the value of \( a^2 + b^2 \) of the given ellipse equation:

\(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\)

1. Recall that the length of the latus rectum of the ellipse is given by \(\frac{2b^2}{a}\). This is given as 10, so we have the equation:
2. Solving for \(b^2\), we multiply both sides by \(a\): \(2b^2 = 10a\)
3. Dividing both sides by 2 gives us: \(b^2 = 5a\)
4. The eccentricity \(e\) of the ellipse is defined as: \(e = \sqrt{1 - \frac{b^2}{a^2}}\)
5. Given the function \(f(t) = t^2 + t + \frac{11}{12}\), we need to find its minimum value. The minimum value of a quadratic function \(at^2 + bt + c\) occurs at \(-\frac{b}{2a}\):
   1. Here, \(a = 1\), \(b = 1\), so the minimum value occurs at \(-\frac{1}{2}\).
   2. Substitute \(t = -\frac{1}{2}\) into \span class="math-tex">f(t): \(f\left(-\frac{1}{2}\right) = \left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right) + \frac{11}{12}\)
   3. Calculate the value: \(=\frac{1}{4} - \frac{1}{2} + \frac{11}{12}\)
   4. Find a common denominator: \(\frac{3}{12} - \frac{6}{12} + \frac{11}{12} = \frac{8}{12} = \frac{2}{3}\)
6. Substitute the value of \(b^2 = 5a\) into the eccentricity equation: \(e = \sqrt{1 - \frac{5a}{a^2}} = \sqrt{1 - \frac{5}{a}}\)
7. Use the fact that the minimum value of \(f(t)\) also equals \(\frac{2}{3}\), the eccentricity must satisfy this constraint.
8. Substitute \(b^2 = 5a\) back into the relation of eccentricity and find the expression for \(a^2 + b^2\):
   1. Since \(b^2 = 5a\), substituting \(b^2\) in terms of \(a\) into \(a^2 + b^2\):
   2. \(a^2 + b^2 = a^2 + 5a\)
9. Finally, match \(a^2 + b^2 = 126\), making this the correct answer.

Thus, the value of \(a^2 + b^2\) is 126.

Answer 2

The equation of the ellipse is given as: \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1. \] The length of the latus rectum \( L \) of the ellipse is given by the formula: \[ L = \frac{2b^2}{a}. \] We are told that the length of the latus rectum is 10, so: \[ \frac{2b^2}{a} = 10 \quad \Rightarrow \quad 5a = b^2. \tag{1} \] Next, we are given the function \( f(t) = t^2 + t + \frac{11}{12} \), and we need to minimize it. The first derivative of \( f(t) \) is: \[ \frac{df(t)}{dt} = 2t + 1. \] Setting the derivative equal to zero to find the critical point: \[ 2t + 1 = 0 \quad \Rightarrow \quad t = -\frac{1}{2}. \] Substitute \( t = -\frac{1}{2} \) into the function \( f(t) \) to find the minimum value: \[ f\left( -\frac{1}{2} \right) = \left( -\frac{1}{2} \right)^2 + \left( -\frac{1}{2} \right) + \frac{11}{12} = \frac{1}{4} - \frac{1}{2} + \frac{11}{12} = \frac{3}{12} - \frac{6}{12} + \frac{11}{12} = \frac{8}{12} = \frac{2}{3}. \] Now, equating this minimum value to the eccentricity: \[ e = \frac{2}{3}. \] Since eccentricity is related to the geometry of the ellipse by: \[ e^2 = 1 - \frac{b^2}{a^2}, \] we substitute \( e^2 = \frac{4}{9} \) and solve for \( \frac{b^2}{a^2} \): \[ \frac{4}{9} = 1 - \frac{b^2}{a^2} \quad \Rightarrow \quad \frac{b^2}{a^2} = \frac{5}{9}. \] Using the relation \( b^2 = 5a \) from equation (1), we substitute this into the above equation: \[ \frac{5a}{a^2} = \frac{5}{9} \quad \Rightarrow \quad \frac{5}{a} = \frac{5}{9} \quad \Rightarrow \quad a = 9. \] Now, substitute \( a = 9 \) into \( b^2 = 5a \): \[ b^2 = 5 \times 9 = 45. \] Finally, calculate \( a^2 + b^2 \): \[ a^2 + b^2 = 9^2 + 45 = 81 + 45 = 126. \] Thus, the correct answer is \( 126 \), which corresponds to option (2).