Question

If the ellipse $\frac{x^2}{4}+y^2=1$ meets the ellipse $x^2+\frac{y^2}{a^2}=1$ at four distinct points and $a=b^2-5b+7$, then b does not lie in

• A. [4, 5]
• B. $(-\infty ,2)\cup (3,\infty )$
• C. $(-\infty ,0)$
• D. [2, 3]

Answer 1

The correct option is D [2, 3]

For the two ellipses to intersect in $4$ distinct points $a>1$.

$\Rightarrow b^2-5b+7>1\Rightarrow b^2-5b+6>0\Rightarrow b\in (-\infty ,2)\cup (3,\infty )$

$\therefore$ $b$ doesn't lie on $[2,3]$.