$A$ chord $PQ$ of the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$ subtends a right angle at its centre. The locus of the point of intersection of tangents drawn at $P$ and $Q$ is-

Lines $x+y=1$ and $3y=x+3$ intersect the ellipse $x^{2}+9y^{2}=9$ at the points $P, Q$ and $R$. The area of the $\triangle PQR$ is

The equation $2x^2 + 3y^2 = 30$ represents

If the chords of contact of tangents drawn from two points $(x_1, y_1)$ and $(x_2, y_2)$ to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ are at right angles,then $\frac{x_1 x_2}{y_1 y_2} = \dots$

The area of the quadrilateral formed by the tangents at the end points of the latus rectum to the ellipse $\frac{x^2}{9} + \frac{y^2}{5} = 1$ is .............. $sq. \text{ units}$.

An ellipse $\frac{(x-x_0)^2}{a^2} + \frac{(y-y_0)^2}{b^2} = 1$ with $a > b$ is tangent to both the $x$ and $y$ axes and is located in the first quadrant. Let $F_1$ and $F_2$ be the two foci of the ellipse and $O$ be the origin such that $OF_1 < OF_2$. Suppose the triangle $OF_1F_2$ is an isosceles triangle with $\angle OF_1F_2 = 120^{\circ}$. Then the eccentricity of the ellipse is: