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Consider the ellipse C:(x^(2))/(a^(2))+(y^(2))/(b^(2))=1 having its centre at the origin O and eccentricity e. Statement-1: If the normal at an end L of a Latusrectum of the ellipse C meets the major axis at G, then OG=ae^(3) Statement-2 : the normal at a point on the ellipse (x^(2))/(a^(2))+(y^(2))/(b^(2))=1 never passes through its foci. |
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Answer» STATEMENT-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1 Then, `alex_(1)leand -bley_(1)ltb`. The equation of the normal at `(x_(1),y_(2))" is "(a^(2)x)/(x_(1))-(b^(2)y)/(y_(1))=a^(2)-b^(2)` It cutsx-axis at `(e^(2)x_(1),0)`. If the normal at `(x_(1),y_(1))` passes through the focus (AE, 0), then `e^(1)x_(1)=aerArrx_(1)=a` Clearly, there is alsopoint on the ellipse WHOSE x-coordinates is more than a. Hence, the normal at any point does not pass through the focus The coordinates of L are `(ae,b^(2)//a)`. Replacing `x_(1)` by ae, we obtain that the normal at L cuts x-axis at `G(ae^(2),0)` `therefore OG=ae^(3)` Thus,both the statements are true and statement-2 is a correct explanation for statement-1. |
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