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.