Show that the maximum distance of centre of the ellipse `x^(2)/a^(2)+y^(2)/b^(2) = 1` from a normal to the ellipse is (a - b).

The maximum distance of the the normal to the ellipse (x^(2))/(9)+(y^(2))/(4)=1 from its centre is

The centre of the ellipse (x+y-2)^(2)/9+(x-y)^(2)/16=1 is

The centre of the ellipse (x+y-3)^(2)/9+(x-y+1)^(2)/16=1 is

The maximum number of normals to hyperbola x^(2)/a^(2)-y^(2)/b^(2)=1 from an external point is

Let d be the perpendicular distance from the centre of the ellipse x^(2)/a^(2)+y^(2)/b^(2) = 1 to the tangent drawn at a point P on ellipse. If F_(1) and F_(2) are the foci of the ellipse, then show that (PF_(1)-PF_(2))^(2)=4a^(2)(1-(b^(2))/(d^(2)))

The pole of line x = a/e with respect to the ellipse x^(2)//a^(2)+y^(2)//b^(2)=1 is

The locus of the foot of the perpendicular drawn from the centre of the ellipse (x^(2))/( a^(2)) +(y^(2))/( b^(2)) =1 to any of its tangents is

Show that the equation of the normal to the ellipse (x^(2))/(a^(2))+(y^(2))/(b^(2))=1 at the positive end of latus rectum is x-ey-e^(3)a=0

The eccentric angles of the extremities of latusrecta of the ellipse x^(2)//a^(2)+y^(2)//b^(2)=1 is