Prove that the product of the perpendiculars from the foci upon any tangent to the ellipse `x^2/a^2+y^2/b^2=1` is `b^2`

The product of the perpendiculars from the foci on any tengent to the ellipse x^2/a^2 + y^2/b^2 = 1 is

The product of the perpendiculars from the foci on any tangent to the ellipse x^(2)//a^(2)+y^(2)//b^(2)=1 is

The product of the perpendiculars from the foci on any tangent to the ellipse (x^(2))/(a^(2)) + (y^(2))/(b^(2)) = 1 is

Prove that the product of the perpendicular from the foci on any tangent to the ellips (x^(2))/(a^(2))+(y^(2))/(b^(2)) =1 is equal to b^(2)

Prove that the product of the perpendicular from the foci on any tangent to the ellips (x^(2))/(a^(2))+(y^(2))/(b^(2)) =1 is equal to b^(2)

Product of perpendiculars drawn from the foci upon any tangent to the ellipse 3x^(2)+4y^(2)=12 is

The product of the perpendicular from the foci on any tangent to the hyperbola x^(2) //a^(2) -y^(2) //b^(2) =1 is

Prove that the product of the distances of the foci from any tangent to the ellipse x^2/a^2+y^2/b^2=1 is equal to b^2 .