Number of tangents to the circle $x^2 + y^2 = 3$ , which are normal to the ellipse $4x^2 + 9y^2 = 36$ , is
Number of tangents to the circle $x^2 + y^2 = 3$ , which are normal to the ellipse $4x^2 + 9y^2 = 36$ , is
- A
$0$
- B
$1$
- C
$2$
- D
$3$
Similar Questions
On the ellipse $\frac{x^{2}}{8}+\frac{y^{2}}{4}=1$ let $P$ be a point in the second quadrant such that the tangent at $\mathrm{P}$ to the ellipse is perpendicular to the line $x+2 y=0$. Let $S$ and $\mathrm{S}^{\prime}$ be the foci of the ellipse and $\mathrm{e}$ be its eccentricity. If $\mathrm{A}$ is the area of the triangle $SPS'$ then, the value of $\left(5-\mathrm{e}^{2}\right) . \mathrm{A}$ is :
On the ellipse $\frac{x^{2}}{8}+\frac{y^{2}}{4}=1$ let $P$ be a point in the second quadrant such that the tangent at $\mathrm{P}$ to the ellipse is perpendicular to the line $x+2 y=0$. Let $S$ and $\mathrm{S}^{\prime}$ be the foci of the ellipse and $\mathrm{e}$ be its eccentricity. If $\mathrm{A}$ is the area of the triangle $SPS'$ then, the value of $\left(5-\mathrm{e}^{2}\right) . \mathrm{A}$ is :
- [JEE MAIN 2021]
Area (in sq. units) of the region outside $\frac{|\mathrm{x}|}{2}+\frac{|\mathrm{y}|}{3}=1$ and inside the ellipse $\frac{\mathrm{x}^{2}}{4}+\frac{\mathrm{y}^{2}}{9}=1$ is
Area (in sq. units) of the region outside $\frac{|\mathrm{x}|}{2}+\frac{|\mathrm{y}|}{3}=1$ and inside the ellipse $\frac{\mathrm{x}^{2}}{4}+\frac{\mathrm{y}^{2}}{9}=1$ is
- [JEE MAIN 2020]
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 equation of the tangents drawn at the ends of the major axis of the ellipse $9{x^2} + 5{y^2} - 30y = 0$, are
The equation of the tangents drawn at the ends of the major axis of the ellipse $9{x^2} + 5{y^2} - 30y = 0$, are
If the normal at one end of the latus rectum of an ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ passes through one end of the minor axis then :
If the normal at one end of the latus rectum of an ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ passes through one end of the minor axis then :