If P is the length of perpendicluar drawn from the origin to any normal to the ellipse `x^(2)/25+y^(2)/16=1`, then the maximum value of p is

If 'd' be the length of perpendicular drawn from origin to any normal of the ellipse (x^(2))/(25)+(y^(2))/(16)=1 then 'd' cannot exceed

If length of perpendicular drawn from origin to any normal of the ellipse (x^(2))/(16)+(y^(2))/(25)=1 is l then l cannot be

The locus of the foot of the perpendicular drawn from the centre on any tangent to the ellipse x^2/25+y^2/16=1 is:

If the normal to the ellipse x^2/25+y^2/1=1 is at a distance p from the origin then the maximum value of p is

The locus of the foot of the perpendicular drawn from the origin to any tangent to the hyperbola (x^(2))/(36)-(y^(2))/(16)=1 is

Let d_(1) and d_(2) be the lengths of perpendiculars drawn from foci S' and S of the ellipse (x^(2))/(a^(2))+(y^(2))/(b^(2))=1 to the tangent at any point P to the ellipse. Then S'P : SP is equal to

Two perpendicular tangents drawn to the ellipse (x^(2))/(25)+(y^(2))/(16)=1 intersect on the curve.

If p is the length of the perpendicular drawn from the origin to the line (x)/(a)+(y)/(b)=1 , then which one of the following is correct?

If p is the length of the perpendicular from the focus S of the ellipse x^(2)/a^(2)+y^(2)/b^(2) = 1 to a tangent at a point P on the ellipse, then (2a)/(SP)-1=