Question

If the ellipse $\frac{x^2}{4}+\frac{y^2}{b^2}=1$ meets the ellipse $\frac{x^2}{1}+\frac{y^2}{a^2}=1$ in four distinct points and $a^2=b^2-4b+8$, then $b$ lies in

• A. $(-\infty ,0)$
• B. $(-\infty ,2)$
• C. $(2,\infty )$
• D. $[2,\infty )$

Answer 1

The correct option is B $(-\infty ,2)$

Given,

$\frac{x^2}{1}+\frac{y^2}{a^2}=1$ -- (i)

$\frac{x^2}{4}+\frac{y^2}{b^2}=1$ -- (ii)

are the two equations of the ellipses
Eliminating $y^2$ from both the equations we get,
$x^2\left(\frac{b^2-4a^2}{4a^2{b}^2}\right)=\frac{b^2-a^2}{a^2{b}^2}$

$\frac{x^2}{4}=\frac{b^2-a^2}{b^2-4a^2}$

Substituting the value of $a^2$ we get,
$\frac{x^2}{4}=\frac{4b-8}{-3b^2+16b-32}$

The denominator is always negative as the discriminant of the expression is negative and the coefficient of $b^2$ is also negative.

Hence, $4b-8<0$
$\Rightarrow b<2$
Eliminating $x^2$ from the two equations we get,

$y^2\left(\frac{4a^2-b^2}{a^2{b}^2}\right)=3$

$\frac{y^2}{3}=\frac{a^2{b}^2}{4a^2-b^2}$

Hence, $4a^2-b^2>0$
$\Rightarrow 3b^2-16b+32>0$.

The discriminant of the expression is less than 0 and the coefficient of $b^2$ is greater than 0. Hence, the inequality holds true for all values of $b$.
Hence the common set of the values of $b$ is $(-\infty ,2)$.

Hence, option B is correct