Question

Let the length of the latus rectum of an ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ $(a>b)$ be $30$. If its eccentricity is the maximum value of the function $f(t)=-\dfrac{3}{4}+2t-t^2$, then $(a^2+b^2)$ is equal to

• A. 276
• B. 516
• C. 256
• D. 496

Hint:

Always convert word problems involving conics into standard formulas before substituting numerical values.

Answer 1

The Correct Option is D

Step 1: Find maximum value of the function.
\[ f(t)=-t^2+2t-\frac{3}{4} \] Maximum occurs at \[ t=\frac{-b}{2a}=1 \] \[ f_{\max}=-1+2-\frac{3}{4}=\frac{1}{4} \] Thus, eccentricity \[ e=\frac{1}{4} \] Step 2: Use latus rectum formula.
For ellipse, \[ \text{Latus rectum}=\frac{2b^2}{a}=30 \Rightarrow b^2=15a \] Step 3: Use eccentricity relation.
\[ e^2=1-\frac{b^2}{a^2} \Rightarrow \frac{1}{16}=1-\frac{15a}{a^2} \] \[ \frac{15}{16}=\frac{15}{a} \Rightarrow a=16 \] Step 4: Compute $a^2+b^2$.
\[ b^2=15a=240 \] \[ a^2+b^2=256+240=496 \]