Consider the ellipse
$\frac{x^2}{4}+\frac{y^2}{3}=1$
Let $H (\alpha, 0), 0<\alpha<2$, be a point. A straight line drawn through $H$ parallel to the $y$-axis crosses the ellipse and its auxiliary circle at points $E$ and $F$ respectively, in the first quadrant. The tangent to the ellipse at the point $E$ intersects the positive $x$-axis at a point $G$. Suppose the straight line joining $F$ and the origin makes an angle $\phi$ with the positive $x$-axis.
$List-I$ | $List-II$ |
If $\phi=\frac{\pi}{4}$, then the area of the triangle $F G H$ is | ($P$) $\frac{(\sqrt{3}-1)^4}{8}$ |
If $\phi=\frac{\pi}{3}$, then the area of the triangle $F G H$ is | ($Q$) $1$ |
If $\phi=\frac{\pi}{6}$, then the area of the triangle $F G H$ is | ($R$) $\frac{3}{4}$ |
If $\phi=\frac{\pi}{12}$, then the area of the triangle $F G H$ is | ($S$) $\frac{1}{2 \sqrt{3}}$ |
($T$) $\frac{3 \sqrt{3}}{2}$ |
The correct option is:
$(I) \rightarrow (R); (II) \rightarrow (S); (III) \rightarrow (Q); (IV) \rightarrow (P)$
$(I) \rightarrow (R); (II) \rightarrow (T); (III) \rightarrow (S); (IV) \rightarrow (P)$
$(I) \rightarrow (Q); (II) \rightarrow (T); (III) \rightarrow (S); (IV) \rightarrow (P)$
$(I) \rightarrow (Q); (II) \rightarrow (S); (III) \rightarrow (Q); (IV) \rightarrow (P)$
Let the length of the latus rectum of an ellipse with its major axis long $x -$ axis and center at the origin, be $8$. If the distance between the foci of this ellipse is equal to the length of the length of its minor axis, then which one of the following points lies on it?
For an ellipse $\frac{{{x^2}}}{9} + \frac{{{y^2}}}{4} = 1$ with vertices $A$ and $ A', $ tangent drawn at the point $P$ in the first quadrant meets the $y-$axis in $Q $ and the chord $ A'P$ meets the $y-$axis in $M.$ If $ 'O' $ is the origin then $OQ^2 - MQ^2$ equals to
Let $PQ$ be a focal chord of the parabola $y^{2}=4 x$ such that it subtends an angle of $\frac{\pi}{2}$ at the point $(3, 0)$. Let the line segment $PQ$ be also a focal chord of the ellipse $E: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1, a^{2}>b^{2}$. If $e$ is the eccentricity of the ellipse $E$, then the value of $\frac{1}{e^{2}}$ is equal to
If $y = mx + c$ is tangent on the ellipse $\frac{{{x^2}}}{9} + \frac{{{y^2}}}{4} = 1$, then the value of $c$ is
Area (in sq. units) of the region outside $\frac{|\mathrm{x}|}{2}+\frac{|\mathrm{y}|}{3}=1$ and inside the ellipse $\frac{\mathrm{x}^{2}}{4}+\frac{\mathrm{y}^{2}}{9}=1$ is