In an ellipse, its foci and ends of its major axis are equally spaced. If the length of its semi-minor axis is $2 \sqrt{2}$, then the length of its semi-major axis is
In an ellipse, its foci and ends of its major axis are equally spaced. If the length of its semi-minor axis is $2 \sqrt{2}$, then the length of its semi-major axis is
- [KVPY 2014]
- A
$4$
- B
$2 \sqrt{3}$
- C
$\sqrt{10}$
- D
$3$
Similar Questions
An ellipse is drawn with major and minor axes of lengths $10 $ and $8$ respectively. Using one focus as centre, a circle is drawn that is tangent to the ellipse, with no part of the circle being outside the ellipse. The radius of the circle is
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Number of common tangents of the ellipse $\frac{{{{\left( {x - 2} \right)}^2}}}{9} + \frac{{{{\left( {y + 2} \right)}^2}}}{4} = 1$ and the circle $x^2 + y^2 -4x + 2y + 4 = 0$ is
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Let $S = 0$ is an ellipse whose vartices are the extremities of minor axis of the ellipse $E:\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1,a > b$ If $S = 0$ passes through the foci of $E$ , then its eccentricity is (considering the eccentricity of $E$ as $e$ )
Let $S = 0$ is an ellipse whose vartices are the extremities of minor axis of the ellipse $E:\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1,a > b$ If $S = 0$ passes through the foci of $E$ , then its eccentricity is (considering the eccentricity of $E$ as $e$ )
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:
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:
- [IIT 2022]
The foci of the ellipse $25{(x + 1)^2} + 9{(y + 2)^2} = 225$ are at
The foci of the ellipse $25{(x + 1)^2} + 9{(y + 2)^2} = 225$ are at