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10-2. Parabola, Ellipse, Hyperbola
easy
If the foci and vertices of an ellipse be $( \pm 1,\;0)$ and $( \pm 2,\;0)$, then the minor axis of the ellipse is
A
$2\sqrt 5 $
B
$2$
C
$4$
D
$2\sqrt 3 $
Solution
(d) $ae = 1$, $a = 2$, $e = \frac{1}{2}$
$b = \sqrt {4\left( {1 – \frac{1}{4}} \right)} = \sqrt 3 $
Hence minor axis $ = 2\sqrt 3 $.
Standard 11
Mathematics
Similar Questions
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:
