UPSC NDA I 2023 Mathematics Solved Paper

Concept:The latus rectum of an ellipse is a chord through a focus perpendicular to the major axis.Its endpoints are $(\pm ae,\ \pm b^{2}/a)$ and the focus itself $(\pm ae,0)$ also lies on it.Explanation:The given ellipse is $x^{2}+4y^{2}=1$.Rewrite as $\frac{x^{2}}{1}+\frac{y^{2}}{1/4}=1$.Thus $a^{2}=1$, $b^{2}=1/4$ so $a=1$, $b=1/2$.Eccentricity $e$ satisfies $a^{2}-b^{2}=a^{2}e^{2}$, i.e. $1-1/4=e^{2}$.So $e^{2}=3/4$ and $e=\sqrt{3}/2$.Foci are $(\pm ae,0)=(\pm\sqrt{3}/2,0)$.Endpoints of latus rectum: $(\pm ae,\ \pm b^{2}/a) = (\pm\sqrt{3}/2,\ \pm 1/4)$.Hence the points $(\sqrt{3}/2,0)$, $(\sqrt{3}/2,1/4)$, and $(\sqrt{3}/2,-1/4)$ all lie on the latus rectum (the vertical line $x=\sqrt{3}/2$ between $y=-1/4$ and $y=1/4$).Answer:All three points 1, 2 and 3 lie on the latus rectum. So the correct option is D (1, 2 and 3).