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Q.

E:x2a2+y2b2=1 and  P:y2=4bx,a>b. Statement 1: The tangent at the positive end of the minor axis of the ellipse. E passes through the positive end of the latus rectum of the parabola P.  Statement 2: If the latus rectum of the parabola P is same as that of the ellipse  E, then eccentricity of  E  is  1/2

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a

STATEMENT- I is True, STATEMENT-2 is True; S TATEMENT-2 is a correct explanation for STATEMENT- I

b

STATEMENT-I is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT- I

c

STATEMENT-I is True, STATEMENT-2 is False

d

STATEMENT-I is False, STATEMENT-2 is True

answer is D.

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Detailed Solution

Tangent at the positive end of the minor axis of  E is y = b which meets the parabola  y2 = 4bx  at the pointb4,b   which is not an end of the latus rectum of  P.  so statement- I is false. In statement-2  if  x = b  and   x = ae represent the same line then   b = ae ⇒ba=e⇒a21−e2=a2e2⇒e2=12⇒e=12.Thus statement-2 is true.
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