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=`

The product of the perpendiculars from the two foci of the ellipse (x^(2))/(9)+(y^(2))/(25)=1 on the tangent at any point on the ellipse

Let d_1a n dd_2 be the length of the perpendiculars drawn from the foci Sa n dS ' of the ellipse (x^2)/(a^2)+(y^2)/(b^2)=1 to the tangent at any point P on the ellipse. Then, S P : S^(prime)P= d_1: d_2 (b) d_2: d_1 d1 2:d2 2 (d) sqrt(d_1):sqrt(d_2)

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

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

If F_(1) and F_(2) are the feet of the perpendiculars from the foci S_(1)andS_(2) of the ellipse (x^(2))/(25)+(y^(2))/(16)=1 on the tangent at any point P on the ellipse,then prove that S_(1)F_(1)+S_(2)F_(2)>=8

The locus of the foot of the perpendicular from the foci an any tangent to the ellipse x^(2)/a^(2) + y^(2)/b^(2) = 1 , 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 the ellipse.If F_(1)&F_(2) are the two foci of the ellipse,then show the (PF_(1)-PF_(2))^(2)=4a^(2)[1-(b^(2))/(d^(2))]

If M_(1) and M_(2) are the feet of perpendiculars from foci F_(1) and F_(2) of the ellipse (x^(2))/(64)+(y^(2))/(25)=1 on the tangent at any point P of the ellipse then