Let $'E'$ be the ellipse $\frac{{{x^2}}}{9}$$+$$\frac{{{y^2}}}{4}$ $= 1$ $\& $ $'C' $ be the circle $x^2 + y^2 = 9.$ Let $P$ $\&$ $Q$ be the points $(1 , 2) $ and $(2, 1)$ respectively. Then :
Let $'E'$ be the ellipse $\frac{{{x^2}}}{9}$$+$$\frac{{{y^2}}}{4}$ $= 1$ $\& $ $'C' $ be the circle $x^2 + y^2 = 9.$ Let $P$ $\&$ $Q$ be the points $(1 , 2) $ and $(2, 1)$ respectively. Then :
- A
$Q$ lies inside $C$ but outside $E$
- B
$Q$ lies outside both $C$ $\&$ $ E$
- C
$P$ lies inside both $C$ $ \&$ $E$
- D
$P$ lies inside $C$ but outside $E.$
Similar Questions
Let $A,B$ and $C$ are three points on ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$where line joing $A \,\,\&\,\, C$ is parallel to the $x-$axis and $B$ is end point of minor axis whose ordinate is positive then maximum area of $\Delta ABC,$ is-
Let $A,B$ and $C$ are three points on ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$where line joing $A \,\,\&\,\, C$ is parallel to the $x-$axis and $B$ is end point of minor axis whose ordinate is positive then maximum area of $\Delta ABC,$ is-
Let $\mathrm{E}$ be an ellipse whose axes are parallel to the co-ordinates axes, having its center at $(3,-4)$, one focus at $(4,-4)$ and one vertex at $(5,-4) .$ If $m x-y=4, m\,>\,0$ is a tangent to the ellipse $\mathrm{E}$, then the value of $5 \mathrm{~m}^{2}$ is equal to $.....$
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- [JEE MAIN 2021]
Latus rectum of ellipse $4{x^2} + 9{y^2} - 8x - 36y + 4 = 0$ is
Latus rectum of ellipse $4{x^2} + 9{y^2} - 8x - 36y + 4 = 0$ 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:
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 angle of intersection of ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ and circle ${x^2} + {y^2} = ab$, is
The angle of intersection of ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ and circle ${x^2} + {y^2} = ab$, is