Question

Let the ellipse \[ E=\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \] have eccentricity equal to the greatest value of the function \[ f(t)=-\frac34+2t-t^2 \] and the length of its latus rectum is $30$. Find the value of $a^2+b^2$.

Hint:

For ellipse problems, always connect eccentricity and latus rectum formulas carefully.

Answer 1

Correct Answer: 496

Step 1: Find the maximum value of the given function.
\[ f(t)=-t^2+2t-\frac34 \] This is a downward opening parabola.
Maximum value occurs at \[ t=\frac{-b}{2a}=\frac{2}{2}=1 \] \[ f_{\max}=-1+2-\frac34=\frac14 \] Step 2: Use eccentricity of ellipse.
Given \[ e^2=\frac14 \Rightarrow e=\frac12 \] For ellipse, \[ e^2=1-\frac{b^2}{a^2} \] \[ \Rightarrow \frac{b^2}{a^2}=\frac{3}{4} \Rightarrow a^2-b^2=\frac{a^2}{4} \] Step 3: Use latus rectum condition.
Length of latus rectum of ellipse is \[ \frac{2b^2}{a}=30 \Rightarrow b^2=15a \quad (1) \] Step 4: Solve using equations.
From eccentricity relation: \[ a^2-b^2=\frac{a^2}{4} \Rightarrow b^2=\frac{3a^2}{4} \quad (2) \] From (1) and (2): \[ 15a=\frac{3a^2}{4} \Rightarrow a^2=256 \Rightarrow a=16 \] \[ b^2=15a=240 \] Step 5: Final calculation.
\[ a^2+b^2=256+240=496 \] Final conclusion.
The required value is 496.