Conic Sections
Conic Section
star_batch_jee_advanced_2025
Grade 11
Question:
If the ellipse $\frac{x^2}{4} + y^2 = 1$ meets the ellipse $x^2 + \frac{y^2}{a^2} = 1$ in four distinct points and $a = b^2 - 5b + 7$, then $b$ does not lie in
$[4, 5]$
$(-\infty, 2) \cup (3, \infty)$
$(-\infty, 0)$
$[2, 3]$
Step-by-Step Solution
Key Concept: Two conics intersect at four distinct points when the discriminant condition satisfies specific inequalities.
For two ellipses to intersect at four distinct points with $a > 1$, the discriminant condition requires $b^2 - 5b + 7 > 1$ and $b^2 - 5b + 6 > 0$. This gives $b^2 - 5b + 7 > 1 \Rightarrow b^2 - 5b + 6 > 0$, which factors as $(b-2)(b-3) > 0$. Therefore $b \in (-\infty, 2) \cup (3, \infty)$, so $b$ does not lie in $[2, 3]$.
Correct Answer: 4