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 :
$Q$ lies inside $C$ but outside $E$
$Q$ lies outside both $C$ $\&$ $ E$
$P$ lies inside both $C$ $ \&$ $E$
$P$ lies inside $C$ but outside $E.$
The eccentricity of the ellipse ${\left( {\frac{{x - 3}}{y}} \right)^2} + {\left( {1 - \frac{4}{y}} \right)^2} = \frac{1}{9}$ is
Find the equation for the ellipse that satisfies the given conditions: Ends of major axis $(±3,\,0)$ ends of minor axis $(0,\,±2)$
A triangle is formed by the tangents at the point $(2,2)$ on the curves $y^2=2 x$ and $x^2+y^2=4 x$, and the line $x+y+2=0$. If $r$ is the radius of its circumcircle, then $r ^2$ is equal to $........$.
Let the line $y=m x$ and the ellipse $2 x^{2}+y^{2}=1$ intersect at a ponit $\mathrm{P}$ in the first quadrant. If the normal to this ellipse at $P$ meets the co-ordinate axes at $\left(-\frac{1}{3 \sqrt{2}}, 0\right)$ and $(0, \beta),$ then $\beta$ is equal to
The equation of the normal at the point $(2, 3)$ on the ellipse $9{x^2} + 16{y^2} = 180$, is