P ( _1 ) and Q ( _2 ) are two points on the ellipse x^2a^2+ y^2b^2=1 with eccentricity e. If PSQ is a focal chord and tan ⁡ ( _12 ) tan ⁡ ( _22 )=- (2 2+3 ), then e and S are

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Correct answer: D. (d) Given equation of ellipse x^2a^2+ y^2b^2=1 , (a>b ) Then, 5= ( ae ,0 ) Slope of PS = Slope of SQ b sin ⁡ _1-0 acos ⁡ _1- ae= bsin ⁡ _2-0 acos ⁡ _2- ae absin ⁡ _1 cos ⁡ _2- abesin ⁡ _1= ab cos ⁡ _1 sin ⁡ _2- abcsin ⁡ _2 ab( sin ⁡ _1 cos ⁡ _2- cos ⁡ _1 sin ⁡ _2)= abc( sin ⁡ _1- sin ⁡ _2) sin ⁡( _1- _2)=e ( sin ⁡ _1 sin ⁡ _2 )e= sin ⁡( _1- _2) sin ⁡ _1- sin ⁡ _2= 2 sin ⁡ ( _1- _22 ) cos ⁡ ( _1- _22 )2 sin ⁡ ( _1- _22 ) cos ⁡ ( _1+ _22 )e= cos ⁡ ( _1- _22 ) cos ⁡ ( _1+ _22 ) Add both side 1 e+1= cos ⁡ ( _1- _22 )+ cos ⁡ ( _1+ _22 ) cos ⁡ ( _1+ _22 )e+1= 2 cos ⁡ ( _1- _2+ _1+ _24 ) cos ⁡ ( _1+ _2- _1+ _24 ) cos ⁡ ( _1+ _22 )e+1= 2 cos ⁡ ( _12 ) cos ⁡ ( _22 ) cos ⁡ ( _1+ _22 ) … … … … (i ) Similarly e-1= cos ⁡ ( _1- _22 )- cos ⁡ ( _1+ _22 ) cos ⁡ ( _1+ _22 ) e-1= 2 sin ⁡ ( _12 ) sin ⁡ ( _22 ) cos ⁡ ( _1+ _22 )……………(ii) Now, Divide (ii)/(i), e-1e+1= tan ⁡ ( _12 ) tan ⁡ ( _22 ) e-1e+1=- (2 2+3 ) e-1=-2 2c-2 2-3e-3 (1+2 2+3 )e=-2 2-3-3e=- (2 2+2 ) (2 2+4 )e=- (2 2+1 ) (2 2+2 ) 2-1 2+1c=- (2-1 )2- 2+2 2-2= -1 2 Eccentricity cannot be negative, then e= -1 2 Here, c= ca ,a= ce a=- 2cc= -a 2 So, focus S (c ,0 )-3s ( -a 2 ,0 ) and c= -1 2

$P (θ_{1}) and Q (θ_{2}) are two points on the ellipse \frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1 with eccentricity e . If PSQ is a focal chord and tan (\frac{θ _{1}}{2}) tan (\frac{θ _{2}}{2}) = - (22 + 3), then e and S are$

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