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JEE MainMathematicsModerateConic Sections

If the line segment joining the foci of the
ellipsex2a2+y2b2\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1(a > b) does not
subtend a right angle at any point on the ellipse, then the range of
eccentricity of the ellipse is;fn nh?kZo`Ùk
x2a2+y2b2\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}}= 1(a > b) s
vUrfjr ugh rks nh?kZo`Ùk

Show answer and solution
∴\therefore No point of intersection of ellipse with the
circle with S_1S_2 as diameter
∴\therefore:i eas S_1S_2 ds lkFk ldZy ds lkkFk nh/kZo`r ds
pkSjkgs sbZ fcanq ugha
⇒\Rightarrow x2a2+y2b2\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} =1 and
rFkk x^2 + y^2 = a^2 - b^2
∴\therefore a^2 - b^2< b^2 ⇒\Rightarrow 2b^2> a^2
⇒\Rightarrow b2a2>12\frac{b^{2}}{a^{2}} > \frac{1}{2} is e
<12< \frac{1}{\sqrt{2}}

Source: Grass-G / M-27 Mathematics Paper 2 15 Jan.docx

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