For the ellipse $\frac{{{x^2}}}{{64}} + \frac{{{y^2}}}{{28}} = 1$, the eccentricity is
$\frac{3}{4}$
$\frac{4}{3}$
$\frac{2}{{\sqrt 7 }}$
$1\over3$
What is the equation of the ellipse with foci $( \pm 2,\;0)$ and eccentricity $ = \frac{1}{2}$
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
Let $L$ be a tangent line to the parabola $y^{2}=4 x-20$ at $(6,2)$ . If $L$ is also a tangent to the ellipse $\frac{ x ^{2}}{2}+\frac{ y ^{2}}{ b }=1,$ then the value of $b$ is equal to ..... .
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:
If the normal at any point $P$ on the ellipse cuts the major and minor axes in $G$ and $g$ respectively and $C$ be the centre of the ellipse, then