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

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

Answer 1

The correct option is C $[2,3]$

For the two ellipses to intersect in $4$ distinct points,
$a^2>1\Rightarrow a>1$ or $a<-1$.
For $a<-1$
$b^2-5b+7<-1\Rightarrow b^2-5b+8<0\Rightarrow (b-5/2{)}^2+\frac{7}{4}<0$
Which is not possible

For $a>1$
$\Rightarrow b^2-5b+7>1$
$\Rightarrow b^2-5b+6>0$
$\Rightarrow b\in (-\infty ,2)\cup (3,\infty )$
$\therefore b$ doesn't lie on $[2,3]$.