If the ellipse x24+y2=1 meets the ellipse x2+y2a2=1 at four points and a=b2−5b+7 then ‘b’ does not lie in

S1≡x24+y21=1S2≡x21+y2a2=1From the diagram, a≥1b2−5b+7≥1⇒b∉2 3

If the ellipse x24+y2=1 meets the ellipse x2+y2a2=1 at four points and a=b2−5b+7 then ‘b’ does not lie in

S1≡x24+y21=1S2≡x21+y2a2=1From the diagram, a≥1b2−5b+7≥1⇒b∉2 3

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

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$(\begin{matrix} - \infty & 0 \end{matrix})$

$[\begin{matrix} 4 & 5 \end{matrix}]$

$(\begin{matrix} - \infty, & 2 \end{matrix}) \cup (3, \infty)$

$[\begin{matrix} 2 & 3 \end{matrix}]$

answer is D.

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$\begin{array}{l} S_{1} \equiv \frac{x^{2}}{4} + \frac{y^{2}}{1} = 1 \\ S_{2} \equiv \frac{x^{2}}{1} + \frac{y^{2}}{a^{2}} = 1 \end{array}$

From the diagram, $a \geq 1$

$b^{2} - 5 b + 7 \geq 1$ $\Rightarrow b \notin [2 3]$

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If the ellipse x24+y2=1 meets the ellipse x2+y2a2=1 at four points and a=b2−5b+7 then ‘b’ does not lie in

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