In the ellipse, minor axis is $8$ and eccentricity is $\frac{{\sqrt 5 }}{3}$. Then major axis is
In the ellipse, minor axis is $8$ and eccentricity is $\frac{{\sqrt 5 }}{3}$. Then major axis is
- A
$6$
- B
$12$
- C
$10$
- D
$16$
Similar Questions
The normal at $\left( {2,\frac{3}{2}} \right)$ to the ellipse, $\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{3} = 1$ touches a parabola, whose equation is
The normal at $\left( {2,\frac{3}{2}} \right)$ to the ellipse, $\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{3} = 1$ touches a parabola, whose equation is
- [AIEEE 2012]
Slope of common tangents of parabola $(x -1)^2 = 4(y -2)$ and ellipse ${\left( {x - 1} \right)^2} + \frac{{{{\left( {y - 2} \right)}^2}}}{2} = 1$ are $m_1$ and $m_2$ ,then $m_1^2 + m_2^2$ is equal to
Slope of common tangents of parabola $(x -1)^2 = 4(y -2)$ and ellipse ${\left( {x - 1} \right)^2} + \frac{{{{\left( {y - 2} \right)}^2}}}{2} = 1$ are $m_1$ and $m_2$ ,then $m_1^2 + m_2^2$ is equal to
Let the eccentricity of an ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1, a>b$, be $\frac{1}{4}$. If this ellipse passes through the point $\left(-4 \sqrt{\frac{2}{5}}, 3\right)$, then $a^{2}+b^{2}$ is equal to
Let the eccentricity of an ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1, a>b$, be $\frac{1}{4}$. If this ellipse passes through the point $\left(-4 \sqrt{\frac{2}{5}}, 3\right)$, then $a^{2}+b^{2}$ is equal to
- [JEE MAIN 2022]
The locus of the middle point of the intercept of the tangents drawn from an external point to the ellipse ${x^2} + 2{y^2} = 2$ between the co-ordinates axes, is
The locus of the middle point of the intercept of the tangents drawn from an external point to the ellipse ${x^2} + 2{y^2} = 2$ between the co-ordinates axes, is
- [IIT 2004]
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]