Mathematics - Ellipse Question with Solution | TestHub

For four distinct intersection points, the ellipses must have different major/minor axes.

Ellipse 1: $\frac{x^2}{4}+\frac{y^2}{1}=1$. Here, $a_1^2=4$, $b_1^2=1$.

Ellipse 2: $\frac{x^2}{1}+\frac{y^2}{a^2}=1$. Here, $a_2^2=1$, $b_2^2=a^2$.

For four distinct points, the ellipses must not be identical, which means $a\ne 1$ and $a\ne 2$. Also, for the second equation to represent an ellipse, $a>0$.

The condition for four distinct intersection points is that the major axis of one ellipse is greater than the major axis of the other, and the minor axis of one is less than the minor axis of the other.

In this case, for the first ellipse, the semi-major axis is 2 and the semi-minor axis is 1.

For the second ellipse, the semi-major axis is $a$ (if $a>1$) or 1 (if $a<1$), and the semi-minor axis is 1 (if $a>1$) or $a$ (if $a<1$).

For four distinct points, we need $a>1$.

Given $a=b^2-10b+25=(b-5{)}^2$.

So, $(b-5{)}^2>1$.

This implies $b-5>1$ or $b-5<-1$.

$b>6$ or $b<4$.

Thus, $b\in (-\infty ,4)\cup (6,\infty )$.

The value of $b$ does not satisfy $b\in [4,6]$.

The final answer is $\boxed{D}$