Ellipse
Latus Rectum and Maximum Eccentricity — Finding $a^2+b^2$
nta_pyq_2026_jan
Grade 11
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
Step-by-Step Solution
Key Concept: $f(t)=-(t-1)^2+\tfrac{1}{4}$. Max value $=\tfrac{1}{4}$, so $e=\tfrac{1}{4}$. $b^2=a^2(1-e^2)=\tfrac{15a^2}{16}$. Latus rectum $=\tfrac{2b^2}{a}=\tfrac{15a}{8}=30\Rightarrow a=16$.
$a=16$, $b^2=240$. $a^2+b^2=496$.
Correct Answer: 4