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]
- A
$(I) \rightarrow (R); (II) \rightarrow (S); (III) \rightarrow (Q); (IV) \rightarrow (P)$
- B
$(I) \rightarrow (R); (II) \rightarrow (T); (III) \rightarrow (S); (IV) \rightarrow (P)$
- C
$(I) \rightarrow (Q); (II) \rightarrow (T); (III) \rightarrow (S); (IV) \rightarrow (P)$
- D
$(I) \rightarrow (Q); (II) \rightarrow (S); (III) \rightarrow (Q); (IV) \rightarrow (P)$
Similar Questions
The centre of the ellipse $4{x^2} + 9{y^2} - 16x - 54y + 61 = 0$ is
The centre of the ellipse $4{x^2} + 9{y^2} - 16x - 54y + 61 = 0$ is
If the tangent at a point on the ellipse $\frac{{{x^2}}}{{27}} + \frac{{{y^2}}}{3} = 1$ meets the coordinate axes at $A$ and $B,$ and $O$ is the origin, then the minimum area (in sq. units) of the triangle $OAB$ is
If the tangent at a point on the ellipse $\frac{{{x^2}}}{{27}} + \frac{{{y^2}}}{3} = 1$ meets the coordinate axes at $A$ and $B,$ and $O$ is the origin, then the minimum area (in sq. units) of the triangle $OAB$ is
- [JEE MAIN 2016]
Minimum distance between two points $P$ and $Q$ on the ellipse $\frac{{{x^2}}}{{25}} + \frac{{{y^2}}}{4} = 1$ , if difference between eccentric angles of $P$ and $Q$ is $\frac{{3\pi }}{2}$ , is
Minimum distance between two points $P$ and $Q$ on the ellipse $\frac{{{x^2}}}{{25}} + \frac{{{y^2}}}{4} = 1$ , if difference between eccentric angles of $P$ and $Q$ is $\frac{{3\pi }}{2}$ , is
The line passing through the extremity $A$ of the major axis and extremity $B$ of the minor axis of the ellipse $x^2+9 y^2=9$ meets its auxiliary circle at the point $M$. Then the area of the triangle with vertices at $A, M$ and the origin $O$ is
The line passing through the extremity $A$ of the major axis and extremity $B$ of the minor axis of the ellipse $x^2+9 y^2=9$ meets its auxiliary circle at the point $M$. Then the area of the triangle with vertices at $A, M$ and the origin $O$ is
- [IIT 2009]
If the maximum distance of normal to the ellipse $\frac{x^2}{4}+\frac{y^2}{b^2}=1, b < 2$, from the origin is $1$ , then the eccentricity of the ellipse is:
If the maximum distance of normal to the ellipse $\frac{x^2}{4}+\frac{y^2}{b^2}=1, b < 2$, from the origin is $1$ , then the eccentricity of the ellipse is:
- [JEE MAIN 2023]