Eccentricity of the ellipse whose latus rectum is equal to the distance between two focus points, is
$\frac{{\sqrt 5 + 1}}{2}$
$\frac{{\sqrt 5 - 1}}{2}$
$\frac{{\sqrt 5 }}{2}$
$\frac{{\sqrt 3 }}{2}$
If $P$ lies in the first quadrant on the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ (where $a > b$ ), and tangent & normal drawn at $P$ meets major axis at the points $T$ & $N$ respectively, then the value of $\frac{{\left( {\left| {{F_2}N} \right| + \left| {{F_1}N} \right|} \right)\left( {\left| {{F_2}T} \right| - \left| {{F_1}T} \right|} \right)}}{{\left( {\left| {{F_2}N} \right| - \left| {{F_1}N} \right|} \right)\left( {\left| {{F_2}T} \right| + \left| {{F_1}T} \right|} \right)}}$ is equal to (where $F_1$ & $F_2$ are the foci $(ae, 0)$ & $(-ae, 0)$ respectively)
The length of the latus rectum of an ellipse is $\frac{1}{3}$ of the major axis. Its eccentricity is
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
The equation of the tangent at the point $(1/4, 1/4)$ of the ellipse $\frac{{{x^2}}}{4} + \frac{{{y^2}}}{{12}} = 1$ is
The position of the point $(1, 3)$ with respect to the ellipse $4{x^2} + 9{y^2} - 16x - 54y + 61 = 0$