ntO is the centre of an ellipse whose semi major axis is b.the ordinate of the point P of the ellipse intersects its auxillary circle at Q(when produced) . The straight through P drawn parallel to OQ cuts the major axis at G .prove then PG=b.n

Q. An ellipse is drawn by taking a diameter of the circle $(x-1{)}^2+y^2=1$ as its semi minor axis and a diameter of the circle $x^2+(y-2{)}^2=4$ as its semi-major axis. If the centre of the ellipse is at the origin and its axes are co ordinate axes then, the ellipse is

Q. An ellipse has its center at $(1,-1)$ and semi- major axis $=$ $8$, which passed through the point $(1,3)$. If the major axis is parallel to $x$-axis then the equation of ellipse is

Q. An ellipse has its centre at $(1,-1)$ and semi major axis equal to $8$. If this ellipse passes through the point $(1,3)$, its eccentricity is equal to

Q. An ellipse is drawn by taking a diameter of the circle $(x-1{)}^2+y^2=1$ as its semi-minor-axis and a diameter of the circle $x^2+(y-2{)}^2=4$ is semi-major axis. If the centre of the ellipse is at origin and its axes are the coordinate axes, then the equation of the ellipse is

Q. An ellipse has its centre at (1,-1) and semi-major axis=8 and it passes through the point (1,3). The equation of the ellipse is