Let $P$ be a variable point on the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ with foci ${F_1}$ and ${F_2}$. If $A$ is the area of the triangle $P{F_1}{F_2}$, then maximum value of $A$ is
$ab$
$abe$
$\frac{e}{{ab}}$
$\frac{{ab}}{e}$
Let $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a>b$ be an ellipse, whose eccentricity is $\frac{1}{\sqrt{2}}$ and the length of the latus rectum is $\sqrt{14}$. Then the square of the eccentricity of $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is :
The distance between the foci of the ellipse $3{x^2} + 4{y^2} = 48$ is
If $x^{2}+9 y^{2}-4 x+3=0, x, y \in R$, then $x$ and $y$ respectively lie in the intervals:
The equation $\frac{{{x^2}}}{{2 - r}} + \frac{{{y^2}}}{{r - 5}} + 1 = 0$ represents an ellipse, if