The ellipse ${x^2} + 4{y^2} = 4$ is inscribed in a rectangle aligned with the coordinate axes, which in trun is inscribed in another ellipse that passes through the point $(4,0) $ . Then the equation of the ellipse is :
$\;{x^2} + 12{y^2} = 16$
$\;4{x^2} + 48{y^2} = 48$
$\;4{x^2} + 64{y^2} = 48$
$\;{x^2} + 16{y^2} = 16$
An ellipse is inscribed in a circle and a point is inside a circle is choosen at random. If the probability that this point lies outside the ellipse is $\frac {2}{3}$ then eccentricity of ellipse is $\frac{{a\sqrt b }}{c}$ . Where $gcd( a, c) = 1$ and $b$ is square free integer ($b$ is not divisible by square of any integer except $1$ ) then $a · b · c$ is
If the normal at the point $P(\theta )$ to the ellipse $\frac{{{x^2}}}{{14}} + \frac{{{y^2}}}{5} = 1$ intersects it again at the point $Q(2\theta )$, then $\cos \theta $ is equal to
The normal at a point $P$ on the ellipse $x^2+4 y^2=16$ meets the $x$-axis at $Q$. If $M$ is the mid point of the line segment $P Q$, then the locus of $M$ intersects the latus rectums of the given ellipse at the points
Find the equation for the ellipse that satisfies the given conditions: Length of major axis $26$ foci $(±5,\,0)$
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