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Range of b for Two Ellipses Meeting in Four Points | JEE

JEE Maths question with a full step-by-step solution.

Question
If the ellipse x24+y2=1\dfrac{x^2}{4}+y^2=1 meets the ellipse x2+y2a2=1x^2+\dfrac{y^2}{a^2}=1 in four distinct points and a=b2−5b+7a=b^2-5b+7, then bb does not lie in:
A[4,5][4,5]
B(−∞,2)∪(3,∞)(-\infty,2)\cup(3,\infty)
C(−∞,0)(-\infty,0)
D[2,3][2,3]correct
Solution
Step 1: Subtract the two equations. From x24+y2=1\dfrac{x^2}{4}+y^2=1 and x2+y2a2=1x^2+\dfrac{y^2}{a^2}=1,
x2(14−1)+y2(1−1a2)=0  ⇒  y2⋅a2−1a2=3x24.x^2\left(\frac14-1\right)+y^2\left(1-\frac{1}{a^2}\right)=0 \;\Rightarrow\; y^2\cdot\frac{a^2-1}{a^2}=\frac{3x^2}{4}.
 Four distinct real intersection points require a2−1>0a^2-1>0, i.e. a>1a>1 (with a>0a>0). Step 2: Substitute a=b2−5b+7a=b^2-5b+7.
b2−5b+7>1  ⇒  b2−5b+6>0  ⇒  (b−2)(b−3)>0.b^2-5b+7>1 \;\Rightarrow\; b^2-5b+6>0 \;\Rightarrow\;(b-2)(b-3)>0.
 Step 3: So b<2b<2 or b>3b>3, i.e. b∈(−∞,2)∪(3,∞)b\in(-\infty,2)\cup(3,\infty). Therefore bb does not lie in [2,3][2,3]. Correct answer: (4)
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