Let $P$ be an arbitrary point on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ $a > b > 0$. Suppose $F_1$ and $F_2$ are the foci of the ellipse. The locus of the centroid of the $\Delta P F_1 F_2$ as $P$ moves on the ellipse is
a circle
a parabola
an ellipse
a hyperbola
The tangent and normal to the ellipse $3x^2 + 5y^2 = 32$ at the point $P(2, 2)$ meet the $x-$ axis at $Q$ and $R,$ respectively. Then the area(in sq. units) of the triangle $PQR$ is
Minimum area of the triangle by any tangent to the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ with the coordinate axes is
The ellipse $E_1: \frac{x^2}{9}+\frac{y^2}{4}=1$ is inscribed in a rectangle $R$ whose sides are parallel to the coordinate axes.
Another ellipse $E _2$ passing through the point $(0,4)$ circumscribes the rectangle $R$.. The eccentricity of the ellipse $E _2$ is
$P$ is a variable point on the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ with $AA'$ as the major axis. Then the maximum value of the area of $\Delta APA'$ is
A rod $AB$ of length $15\,cm$ rests in between two coordinate axes in such a way that the end point A lies on $x-$ axis and end point $B$ lies on $y-$ axis. A point $P(x,\, y)$ is taken on the rod in such a way that $AP =6\, cm .$ Show that the locus of $P$ is an ellipse.